Question

A projectile is launched from the top of a 75m-tall cliff with an initial velocity of 31m/s at an angle of 42° above the horizontal. For this problem, assume that the base of the cliff is at y 0 and that the projectile starts at x 0 on a standard xy plane 1. What is the maximum height above the ground that the projectile will achieve? 2. How long will it take for the projectile to reach its maximum height? 3. How long will it take for the projectile to reach the ground? Note: you are being asked to find the total time from when the projectile was initially launched to when it hits the ground, not from the highest point. 4 How far from the base of the cliff will the projectile be when it hits the ground? 5. How fast will the projectile be moving when it hits the ground? Note: remember that the ball will be moving along both the x- and y-axes at this point!

Answer 1

a)

max height

H = h + (u sin x) ^2/ 2g = 75 + ( 31 sin 42)^2 / 19.6

H =

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b)

t = u sin x /g = 31 sin 42 / 9.8 = 2.117 s

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c)

using 2nd equation of motion

h = ho + u sin x.* t - 0.5 gt^2

0 = 75 + 31 sin 42 * t - 4.9 t^2

t = 6.565 s

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d)

R = 31 cos 42 * 6.565 = 151.24 m

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e)

using conservation of energy

0.5 m v^2 = 0.5 m u^2 + mgh

0.5 v^2 = 0.5* 31^2 + 9.8* 75

v = 49.305 m/s

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