19 October, 18:18

# 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?

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1. 19 October, 18:35
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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
2. 19 October, 20:47
0
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.