Projectile Motion Questions Help (1 Viewer)

[shantu1992]

A projectile is fired from the top of a cliff with a velocity of 80 ms-1 at an angle of 30 degrees to the horizontal. The cliff is 100 m above sea level.

ignoring air resistance calculate;

(A) the time taken for the projectile to reach its maximum height
(b) the maximum height above sea level reached by the projectile
(c) the speed of the projectile at its maximum heigh
(d) the horizontal distance from teh base of the cliff that the projectile enters the sea

 

[SkimDawg]

a) 4 seconds
b) 180 metres
d) 692.8 metres

b)
Uy = 80sin30
= 40m/s
ay = -10m/s(-2)
Delta(y) = ?
V^2y = U^2y + 2ayDelta(y)
0 = 40^2 + (2 x -10Delta(y))
20Delta(y) = 1600
Delta(y) = 80 + 100 (height of cliff)
= 180m
a)
Vy = Uy + at
0 = 40 + -10t
t = 4 seconds

d)
Delta(y) = Uyt + 1/2at^2
- 180 = + 1/2(-10)t^2
t = sqrt(-180 x 2/-10)
= 6 seconds
Therefore total time = 4 + 6 = 10 seconds
Now Delta(x) = Uxt
Ux = 80cos30
= 69.28m/s
Delta(x) = Uxt
= 69.28 x 10
= 692.8 metres

Sorry about the order, it seemed easier figuring it out as a) needed the value from b).
For c), i think its 0, im not sure though.

 

[clintmyster]

at first glance part c) looks like 0 because at the maximum height the projectille should be stationary

 

[smp211]

Yes, A B and D are correct.

C, however, is wrong.

At the maximum height, vertical velocity is 0, due to the acceleration of gravity downards. However, we know that horizontal velocity remains constant throughout the flight.

Therefore, the answer to C is the horizontal velocity, 69.28 ms^-1 due right.

 

[cutemouse]

A projectile is fired from the top of a cliff with a velocity of 80 ms-1 at an angle of 30 degrees to the horizontal. The cliff is 100 m above sea level.

ignoring air resistance calculate;

(A) the time taken for the projectile to reach its maximum height

t_up=(usinθ)g
=(80*sin30)/9.8
=4.08s

(b) the maximum height above sea level reached by the projectile

h=(u²sin²θ)/2g
=(80²sin²30)/(2*9.8)
=81.63m

Therefore height above sea level = 181.63m

(c) the speed of the projectile at its maximum heigh

ie when V_y=0
Therefore u_x=69.28ms^-1

(d) the horizontal distance from teh base of the cliff that the projectile enters the sea

Δy=u_yt+(1/2)a_yt²
181.63=(0)t+(1/2)*9.8*t²
t=6.09
Δx₁=u_xt
=69.28*6.09
=421.92m

Δx₂=69.28*4.08
=282.66

Therefore projectile lands 704.58m from cliff's base.

Hope that helps,

Jason

 

[cutemouse]

Oh btw, you probably wont get a question like this in the actual HSC, as it's considered to be too hard (because it involves both full flight and half flight projectile motion). But still it's good practise.

Regards,

Jason