Question

5. A cricket ball is thrown at a speed of 28m/s in a direction 30° above the horizontal Calculate (a) the
maximum height (b) the time taken by the ball to return the same level. And (c) the horizontal distance
from the thrower to the point where the ball returns to the same level.​

Answer 1

Given:

• Initial velocity(u)= 28 m/s
• Angle of projection (Ф)= 30°

Solution:

(a) Maximum height(H)

⇒ H = u² sin²Ф/ 2g

⇒ H = 28 x 28 x sin² 30 / 2 x 10

⇒ H = 28 x 28 x 1 / 4 x 2 x 10

⇒ H = 784/80

⇒ H = 9.8 m

The  maximum height is 9.8 m.

(b) Time of flight(T)

⇒ T = 2 u sinФ/g

⇒ T = 2 x 28 x sin 30 / 10

⇒ T = 2 x 28 x 1 / 2 x 10

⇒ T = 56/20

⇒ T = 2.8s

The time taken by the ball to return the same level is 2.8 s.

(c) Horizontal Range (R)

⇒ R = u² sin2Ф/ g

⇒ R = 28 x 28 x sin2(30)/ 10

⇒ R = 784 x sin60 / 10

⇒ R = 784 x √3 / 2 x 10

⇒ R = 784 x 1.73 / 20

⇒ R = 67.8 m

The horizontal distance  from the thrower to the point where the ball returns to the same level is 67.8 m.