A ball is thrown upward at an angle of 30° to the horizontal and lands on the top edge of a building that is 20 m away. The top edge is 5.0 m above the throwing point. How fast was the ball thrown? Use: sin30°=0.5 and cos30°=0.9 a. 20 m/s
b. 11 m/s
c. 52.3 m/s
d. 16 m/s

The correct option is;

a. 20 m/s

Step-by-step explanation:

The given parameters are;

The angle at which the ball is thrown, θ = 30° to the horizontal

The horizontal distance of the top edge of the building where the ball lands from where the ball is thrown, x = 20 m

The height of the top edge of the building above the throwing point = 5 meters

Let "v" represent the speed with which the ball is thrown

We have;

The vertical component of the speed with which the ball is thrown, = v × sin(θ) = v × sin(30°) = v × 0.5 = 0.5·v

= 0.5·v

The horizontal component of the speed with which the ball is thrown, vₓ = v × cos(θ) = v × cos(30°) = v × 0.9 = 0.9·v

vₓ = 0.9·v

The kinematic equation of the motion is y = ·t - (1/2)·g·t², where;

y = The vertical height reached = 5 metes

t = The time taken to reach the specified 5 m, height

g = The acceleration due to gravity = 9.8 m/s², we have;

Therefore, we have;

5 = 0.5·v·t - (1/2)·9.8·t²...(1)

Also, from the horizontal motion of the ball, we have the following kinematic equation of motion;

x = vₓ × t

Therefore, by substituting the known values, we have;

20 = 0.9·v × t

∴ v = 20/(0.9·t) = 200/(9·t)...(2)

Substituting the value of t in equation (1) gives;

5 = 0.5·v·t - (1/2)·9.8·t² = 0.5·(200/(9·t))·t - (1/2)·9.8·t²

∴ 5 = 0.5·(200/(9·t))·t - (1/2)·9.8·t² = 100/9 - 4.9·t²

4.9·t² = 100/9 - 5 = 55/9

t = √(55/(9 × 4.9)) ≈ 1.116766

The time taken to reach the specified 5 m height = t ≈ 1.116766 seconds

From equation (2), we have, v = 200/(9·t) = 200/(9 × 1.116766) ≈ 19.8987 m/s

The speed with which the ball is thrown = v ≈ 19.8987 m/s ≈ 20 m/s. to the nearest whole number.

The speed with which the ball is thrown is approximately 20 m/s

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