# I am self-learning mechanics (projectile motion). How do I solve the question below? (1 Viewer)

#### catha230

##### Member
Hi all, I have just learnt the basic of 4u mechanics and I feel quite overwhelmed by the number of formulae I have to remember for this topic. I also don't know which formulae I should use. I am getting stuck on the question below, I got up to the expression linking t and alpha but since I don't have t, I don't know what to do next. Could someone please help me out with this question? Thank you!

#### waitasecond

##### New Member
hello! try getting the equation: y=-10^2+vsintheta+2
then sub in t=x/(25costheta)
then sub in x= 20, y = 15
then solve for the quadratic in terms of tan theta. u will get 2 values, its between those
hope that helped

#### 5uMath

##### Member
Hi all, I have just learnt the basic of 4u mechanics and I feel quite overwhelmed by the number of formulae I have to remember for this topic. I also don't know which formulae I should use. I am getting stuck on the question below, I got up to the expression linking t and alpha but since I don't have t, I don't know what to do next. Could someone please help me out with this question? Thank you!
View attachment 28224
I recommend you dont memorise formulae, instead learn how to deduce them. Nearly every year projectiles you are asked to first prove a formula then use it.

As for the question above, use horizontal range to find an expression for range and solve for horizontal range > 20, then test for which value the projectile exceeds the building.

#### Drongoski

##### Well-Known Member
But do we know how wide the "rectangular" roof top is? May overshoot the "flat" roof if not very wide, unless question wants us to hit the near tip of roof.

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#### 5uMath

##### Member
But do we know how wide the "rectangular" roof top is? May overshoot the "flat" roof if not very wide, unless question wants us to hit the near tip of roof.
Using the method I suggested, you would get alpha > some real value. Then you would notice the vector from the mans top to the building top has a magnitude to be determined, height of building from mans perspective is 13 (15 - height of man) with horizontal path 20. Call top of building width x, then 20 + x is length of horizontal inclusive of building. You would notice alpha > angle of inclination from man to very end of the top of the building. Therefore call theta this new angle, apply trig and agebra, find min x (find expression for theta using trig ratios) with inequality alpha > new angle, and you have x. Sub x into your expression for theta, then you have minimum. Quite long, but an interesting approach. Btw, some are greater than or equal to, not just >.

Edit, I didnt mention that the vector makes right angle triangle so thats where you would get trig ratios.

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#### CM_Tutor

##### Well-Known Member
But do we know how wide the "rectangular" roof top is? May overshoot the "flat" roof if not very wide, unless question wants us to hit the near tip of roof.
You are right, we don't.

Take the point at the top of the building and closest to the point of projection as $\bg_white T$. Taking the projection as from $\bg_white M(0, 2)$, then $\bg_white T$ is $\bg_white (20, 15)$. If we require the projectile to pass through $\bg_white T$, we will get a quadratic equation in $\bg_white \tan \theta$ that yields two angles, $\bg_white \alpha_1$ and $\bg_white \alpha_2$ where $\bg_white \alpha_1 > \alpha_2$. Firing the projectile at $\bg_white \theta = \alpha_1$ corresponds to the given diagram, with the projectile rising to a maximum height and then passing through $\bg_white T$ on its descent. Firing at $\bg_white \theta = \alpha_2$ corresponds to the projectile passing through $\bg_white T$ as it ascends and then continuing on to reach its maximum height. For any angle $\bg_white \alpha_1 > \theta > \alpha_2$, when the projectile has travelled a distance of $\bg_white x = 20$ m horizontally, it will have a height $\bg_white h$ such that it is some distance above the leading edge of the building, $\bg_white h = y - 15 > 0$.

If we assume that the depth / width of the building in our horizontal direction is great relative to the scale of the motion, any angle $\bg_white \alpha_1 > \theta > \alpha_2$ will result in a roof landing. If it is not, then you are correct, the angle needs to be further constrained to ensure a roof landing, but we also need further information to find a solution.

If you think it is appropriate, you could examine what is the maximum possible range for a landing on the roof and then consider whether this is reasonable for a roof size. You could also examine whether the angles near $\bg_white \alpha_1 > \theta$ or those near $\bg_white \theta > \alpha_2$ will land closest to $\bg_white T$ and thus would be safer choices for a roof landing.

#### CM_Tutor

##### Well-Known Member
Note: These latter considerations arose in an HSC some years ago when the projectile question concerned a cricket player hitting a six that was required to pass over the boundary fence and then land as close as possible to its base.

#### 5uMath

##### Member
You are right, we don't.

Take the point at the top of the building and closest to the point of projection as $\bg_white T$. Taking the projection as from $\bg_white M(0, 2)$, then $\bg_white T$ is $\bg_white (20, 15)$. If we require the projectile to pass through $\bg_white T$, we will get a quadratic equation in $\bg_white \tan \theta$ that yields two angles, $\bg_white \alpha_1$ and $\bg_white \alpha_2$ where $\bg_white \alpha_1 > \alpha_2$. Firing the projectile at $\bg_white \theta = \alpha_1$ corresponds to the given diagram, with the projectile rising to a maximum height and then passing through $\bg_white T$ on its descent. Firing at $\bg_white \theta = \alpha_2$ corresponds to the projectile passing through $\bg_white T$ as it ascends and then continuing on to reach its maximum height. For any angle $\bg_white \alpha_1 > \theta > \alpha_2$, when the projectile has travelled a distance of $\bg_white x = 20$ m horizontally, it will have a height $\bg_white h$ such that it is some distance above the leading edge of the building, $\bg_white h = y - 15 > 0$.

If we assume that the depth / width of the building in our horizontal direction is great relative to the scale of the motion, any angle $\bg_white \alpha_1 > \theta > \alpha_2$ will result in a roof landing. If it is not, then you are correct, the angle needs to be further constrained to ensure a roof landing, but we also need further information to find a solution.

If you think it is appropriate, you could examine what is the maximum possible range for a landing on the roof and then consider whether this is reasonable for a roof size. You could also examine whether the angles near $\bg_white \alpha_1 > \theta$ or those near $\bg_white \theta > \alpha_2$ will land closest to $\bg_white T$ and thus would be safer choices for a roof landing.
Yes I corrected myself earlier. Have you seen the method to use if you were to take this approach that I described?

#### CM_Tutor

##### Well-Known Member
Yes I corrected myself earlier. Have you seen the method to use if you were to take this approach that I described?
I don't recall seeing a question to land a projectile within a horizontal range at an elevated height, though I have seen a question with a target range at the same height as projection (a rotating sprinkler projecting water within a range $\bg_white \alpha_1 \leqslant \theta \leqslant \alpha_2$).

Thinking about the building, it is clear to me that reducing the angle of projection below $\bg_white \alpha_1$ will increase the range and thus move the point of landing along the roof until a new limit is reached (call it $\bg_white \theta = \alpha_3$) that corresponds to landing at $\bg_white (x + 20, 15)$. We also know that if $\bg_white \theta < \alpha_2$, the projectile hits the building, that $\bg_white \theta = \alpha_2$ is a solution, but for a slight increase in $\bg_white \theta$, the path could well continue over the building completely... or it could still land (if the max height was 15.1 m, say).

We will clearly have a solution range $\bg_white \alpha_1 \geqslant \theta \geqslant \alpha_3$ and a solution at $\bg_white \theta = \alpha_2$, but are there / must there be some further solutions on $\bg_white (\alpha_1, \alpha_2)$, and do their existence or nature depend on the width of the building? Without sitting and trying to solve it algebraically, I'm not immediately sure.

#### 5uMath

##### Member
I don't recall seeing a question to land a projectile within a horizontal range at an elevated height, though I have seen a question with a target range at the same height as projection (a rotating sprinkler projecting water within a range $\bg_white \alpha_1 \leqslant \theta \leqslant \alpha_2$).

Thinking about the building, it is clear to me that reducing the angle of projection below $\bg_white \alpha_1$ will increase the range and thus move the point of landing along the roof until a new limit is reached (call it $\bg_white \theta = \alpha_3$) that corresponds to landing at $\bg_white (x + 20, 15)$. We also know that if $\bg_white \theta < \alpha_2$, the projectile hits the building, that $\bg_white \theta = \alpha_2$ is a solution, but for a slight increase in $\bg_white \theta$, the path could well continue over the building completely... or it could still land (if the max height was 15.1 m, say).

We will clearly have a solution range $\bg_white \alpha_1 \geqslant \theta \geqslant \alpha_3$ and a solution at $\bg_white \theta = \alpha_2$, but are there / must there be some further solutions on $\bg_white (\alpha_1, \alpha_2)$, and do their existence or nature depend on the width of the building? Without sitting and trying to solve it algebraically, I'm not immediately sure.
You then would use the horizontal displacement formula x =vcos(alpha),apply the same process except with this formula.

Minimum achieved by simple trig by referring to the building as a side of a right triangle, the base 20 and height 13 (15-2). Tan alpha = 13/20 so alpha = 33 is a minimum

Using approximate values and minimums the answer can be anywhere from 32 <= alpha <= 33, i know this is not the absolute range and is probably flawed. Ill have a look at this again when im free

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#### CM_Tutor

##### Well-Known Member
PS and FYI: I have seen a question like this in a past trial paper where the projectile was a glob of slime (or some such) that hit the side of the building and then slid down from the impact point under gravity and with a resistance proportional to $\bg_white v$ or $\bg_white v^2$ (I don't recall which) and the speed at which it returned to ground level, as a percentage of the velocity of projection, was sought.