Question

A ball is launched from the ground toward a tall wall of a building at initial speed v₀ = 24 m/s at an angle of 39 degrees relative to the horizontal. The wall is at a distance of 32 m from the launch point. How far above the launch point does the ball hit the wall?

Your answer should be in meters, but enter only the numerical part in the answer box.

Answer 1

In projectile motion, range is given by:

R = V0x*T

V0x = V0*cos A = 24*cos 39 deg = 18.65 m/sec

R = range of ball = 32 m

T = time taken by ball to reach the wall = R/V0x

T = 32/18.65 = 1.716 sec

Now Using this value in 2nd kinematic equation in vertical direction:

h = V0y*T + (1/2)*ay*T^2

V0y = Initial vertical velocity = 24*sin 39 deg = 15.10 m/sec

ay = acceleration = -g = -9.81 m/sec^2

So,

h = 15.10*1.716 + (1/2)*(-9.81)*1.716^2

h = 11.5 m = height of ball when it hits the wall

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