A golf ball is driven from a point A with a speed of 40 m/s at an angle of elevation of 30°. On its downward flight, the ball hits an advertising hoarding at a height 15.1 m above the level of A, as shown in the diagram below. Find A. The time taken by the ball to reach its greatest height above A, B. The time taken by the ball to travel from A to B, C. The speed with which the ball hits the hoarding. 20

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20-07-2022
Physics

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A golf ball is driven from a point A with a speed of 40 m/s at an angle of elevation of 30°. On its downward flight, the ball hits an advertising hoarding at a height 15.1 m above the level of A, as shown in the diagram below. Find A. The time taken by the ball to reach its greatest height above A, B. The time taken by the ball to travel from A to B, C. The speed with which the ball hits the hoarding. 20

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onyebuchinnaji onyebuchinnaji · Answer 1 · 2022-07-20T22:54:03+02:00

(a) The time taken by the ball to reach its greatest height is 2 seconds.

(b) The time taken by the ball to travel from A to B, C is 4.08 m/s.

(c) The speed with which the ball hits the hoarding is 69.3 m/s.

Maximum height of the projectile

H = u²sin²θ/2g

H = (40² x (sin 30)²) / (2 x 9.8)

H = 20.4 m

Time for the ball to travel 20.4 m

h = ut - ¹/₂gt²

20.4 = (40 x sin 30)t - (0.5)(9.8)t²

20.4 = 20t - 4.9t²

4.9t² - 20t + 20.4 = 0

solve the quadratic equation using formula method;

t = 2 seconds

Time taken for the ball to travel A, B , C

This is the time of motion of the ball;

T = 2usinθ/g

T = (2 x 40 sin 30)/9.8

T = 4.08 s

Speed of the ball when it hits the hoarding

v = u + gt

vy = 40 sin30 + (9.8 x 4.08)

vy = 59.98 ms

vx = 40 x cos30

vx = 34.64 m/s

vf = √(vy² + vx²)

vf = √(59.98² + 34.64²)

vf = 69.3 m/s

Thus, the time taken by the ball to reach its greatest height is 2 seconds.

The time taken by the ball to travel from A to B, C is 4.08 m/s.

The speed with which the ball hits the hoarding is 69.3 m/s.

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