Question

A cannon, elevated at 40∘ is fired at a wall 300 m away on level ground, as shown in the figure below. The initial speed of the cannonball is 89 m/s. At what height h does the ball hit the wall?

Answer 1

First, we calculate the horizontal and vertical components of the firing velocity.
Vx = 89cos(40)
Vx = 68.2 m/s
Vy = 89sin(40)
Vy = 57.2 m/s
The time of flight for the cannon ball can be obtained by dividing its horizontal distance covered by horizontal velocity.
t = 300/68.2
t = 4.40 s
Using 
s = ut + 0.5at²
for the vertical direction
s = 57.2 × 4.40 + (-9.81)(4.40)²
s = 61.8 m is the height at which the cannon ball hits the wall

Answer 2

Vo = 89 m/s
angle: 40°

=> Vox = Vo * cos 40° = 89 * cos 40°

=> Voy = Vo. sin 40° = 89 * sin 40°

x-movement: uniform => x =Vox * t = 89*cos(40)*t

x = 300 m => t = 300m / [89m/s*cos(40) = 4.4 s

y-movement: uniformly accelerated => y = Voy * t - g*t^2 /2

y = 89m/s * sin(40) * (4.4s) - 9.m/s^2 * (4.4)^2 / 2 = 156.9 m = height the ball hits the wall.