Projectile questions (1 Viewer)

[davidbarnes]

Its late and I'm tired and I just can't seem to solve these 2 problems and get the books asnwers (listed below). So can somone pelase explain how to solve these, etc as I'm doing something wrong or overlooking something.

1. "A volley ball player sets the ball for a team mate. In doing so she taps the ball up at 5.0 m s^-1 at an angle of 80.0 degrees above the horizontal. If her fingers tapped the ball at a height of 1.9 m above the floor, to what maximum height does the ball rise?" Answer = 3.14 m

2. "An 'extreme' cyclist wants to perform a stunt in which he rides up a ramp and launchign himself into the air, then flies throguh a hoop and lands on another ramp. The angle of each ramp is 30.0 degrees, and the cyclist is able to reach the launch height of 1.50 m with a launching speed of 30.0 kph^-1.
(a) At what maxiumum height above the ground could the lower edge of the hoop be placed?
(b) How far away should the landing ramp be placed?"
Asnwers: (a) = 2.39 m (b) = 6.14 m.

Thanks. May Galileo be with you...

 

[watatank]

Its late and I'm tired and I just can't seem to solve these 2 problems and get the books asnwers (listed below). So can somone pelase explain how to solve these, etc as I'm doing something wrong or overlooking something.

1. "A volley ball player sets the ball for a team mate. In doing so she taps the ball up at 5.0 m s^-1 at an angle of 80.0 degrees above the horizontal. If her fingers tapped the ball at a height of 1.9 m above the floor, to what maximum height does the ball rise?" Answer = 3.14 m

2. "An 'extreme' cyclist wants to perform a stunt in which he rides up a ramp and launchign himself into the air, then flies throguh a hoop and lands on another ramp. The angle of each ramp is 30.0 degrees, and the cyclist is able to reach the launch height of 1.50 m with a launching speed of 30.0 kph^-1.
(a) At what maxiumum height above the ground could the lower edge of the hoop be placed?
(b) How far away should the landing ramp be placed?"
Asnwers: (a) = 2.39 m (b) = 6.14 m.

Thanks. May Galileo be with you...

first one:

initial vertical velocity = vsin@ = 5 * sin(80)

final vertical velocity = 0

a = 9.8

v² = u² + 2as

s = (v² - u²)/2a, sub values in

don't forget to add the height above the floor, i did the height above the hand.

second one is trickier, shall do that one when i'm less tired.

 

[Forbidden.]

Its late and I'm tired and I just can't seem to solve these 2 problems and get the books asnwers (listed below). So can somone pelase explain how to solve these, etc as I'm doing something wrong or overlooking something.

1. "A volley ball player sets the ball for a team mate. In doing so she taps the ball up at 5.0 m s^-1 at an angle of 80.0 degrees above the horizontal. If her fingers tapped the ball at a height of 1.9 m above the floor, to what maximum height does the ball rise?" Answer = 3.14 m

2. "An 'extreme' cyclist wants to perform a stunt in which he rides up a ramp and launchign himself into the air, then flies throguh a hoop and lands on another ramp. The angle of each ramp is 30.0 degrees, and the cyclist is able to reach the launch height of 1.50 m with a launching speed of 30.0 kph^-1.
(a) At what maxiumum height above the ground could the lower edge of the hoop be placed?
(b) How far away should the landing ramp be placed?"
Asnwers: (a) = 2.39 m (b) = 6.14 m.

Thanks. May Galileo be with you...

2.
a)


a_y = - 9.8 ms^-2
(Gravity is always constantly acting downwards)

u = 30 / 3.6 = 8.33 ms^-1
(3.6 kmh^-1 = 1 ms^-1)

v_y² - u_y² = 2a_yΔy
v_y² = 0 at maximum height

0 - u_y² = 2a_yΔy
Δy = - u_y² / 2a_y
Δy = - (8.33sin30)² / 2 (-9.8)
Δy = 0.89m

But there is an additional 1.50 m to the launch height in addition to the cyclist's own launch height.
(The answer would be 0.89m if the launch height was 0 m instead of 1.50 m)

0.89m + 1.5m = 2.39m


2.
b)


a_y = - 9.8 ms^-2

u_y = 8.33cos30 = 7.21 ms^-1

Δy = u_yt + ½a_yt²
Δy = 0 when the cyclist touches down on the ground at the very egde of the ramp

0 = (4.17)t + ½(-9.8)t²
0 = 4.17t - 4.9t²
4.9t² = 4.17t
4.9t = 4.17
t = 4.17 / 4.9
t = 0.85 s

Δx = u_xt
Δx = (8.33cos30)(0.85)
Δx = 6.13m

(I used only 2 decimal places throughout the entire working out and answers for simplicity in showing working out, but it is good practice to use full calculator display when working out, while writing out your rounded answer.
This is why I have 6.13m as my answer.)

The attachment below should assist your understanding of the question, working out and answers for question 2.

 

[davidbarnes]

Thankyou Forbidden that is great, and that diagram certainly helps a lot.

 

[davidbarnes]

Heres a really stupid question. If a projectile is dropped from a plane flying 140 m s^-1, what is its inital velocity. Is it 0 m s^-1 or is it 140 m s^-1?

 

[watatank]

Heres a really stupid question. If a projectile is dropped from a plane flying 140 m s^-1, what is its inital velocity. Is it 0 m s^-1 or is it 140 m s^-1?

0 m/s is the initial vertical velocity, and 140m/s is initial horizontal (stays constant throughout)