Projectile motion analyzed?

Question

A stone is projected at a cliff of height h w/ an initial speed of 42.0 m/s directed at angle theta 0=60.0 degrees above the horizontal. The stone strikes at A, 5.50s after launching. Find (a) the height h of the cliff, (b) the speed of the stone just before impact of A, and (c) the maximum height H reached above the ground.

Answer 1

For this problem you only need the y-component of the initial velocity, which is just

Vyo = 42 * cos (60) = 36.4 m/s

(a) To find the height of the cliff, just determine the y-position of the projectile at t=5.5 sec:

Hcliff = Voy * t + 1/2 * g * t^2

Hcliff = 36.4 * 5.5 + 1/2 * (-9.8) * 5.5^2

Hcliff = 52 m

(b) Assume no air resistance, so the x-component of velocity is constant:

Vx = 42 * cos (60) = 21 m/s

The y-component can be found from:

Vy = Vyo + g * t

Vy = 36.4 + (-9.8 * 5.5)

Vy = -17.5 m/s (down)

To find the magnitute of the velocity just use

V = SQRT (Vx^2 + Vy^2)

V = SQRT [21^2 + (-17.5)^2]

V = 27.3 m/s

(c) The maximum height is achieved when Vy = 0, so to find the Hmax use:

Vf^2 = Vo^2 + 2 * g * Hmax; substituding you get

0 = 36.4^2 + 2 * (-9.8) * Hmax; solving:

Hmax = 67.6 m

Good Luck!