Question

-A cricket ball is thrown at a speed of 49 ms – 1

in a direction 30° above the

horizontal. Calculate (5)

(a) the maximum height.

(b) the time taken by the ball to return to the same level.

(c) the distance from the thrower to the point where the ball returns to the same

level.​

Answer 1

QUESTION :

A cricket ball is thrown at a speed of 49 m/s in a direction 30° above the horizontal. Calculate the following :-

(a) the maximum height.

(b) the time taken by the ball to return to the same level.

(c) the distance from the thrower to the point where the ball returns to the same level.

SOLUTION :

Given : u=49 m/s θ=30°

where :-

• u= initial velocity
• θ = degree of direction above the horizontal

(a) maximum height :-

$Maximum \: height \: H= \frac{ {u}^{2} {(sin \:θ )}^{2} }{2g}$

now put :

• u = 49
• θ = 30°
• sinθ = sin 30° = 1/2 or 0.5
• g = 10

$H= \frac{ {(49)}^{2} {(0.5 )}^{2} }{2 \times 10}$

$H= \frac{ 49 \times 49 \times 0.5 \times 0.5 }{2 \times 10}$

$H= \frac{ 49 \times 49 \times 5 \times 5 }{2 \times 1000}$

$H= \frac{60025 }{2 \times 1000}$

$H= \frac{30012.5 }{1000}$

$H= {30.0125 }m$

(b) time taken by the ball to return to the same level:-

$T \: = \frac{2u \: sin \:θ }{g}$

where :

• u = 49
• θ = 30°
• sinθ = sin 30° = 1/2 or 0.5
• g = 10
• T = time taken

$T \: = \frac{2 \times 49 \times \: 0.5 }{10}$

$T \: = \frac{2 \times 49 \times \: 5 }{100}$

$T \: = \frac{ 490}{100}$

$T \: = 4.9s$

(c) distance from the thrower to the point where the ball returns to the same level ( i.e the horizontal range ) :-

$</p><p>R = ucosθ×T$

where :

• u = 49
• θ = 30°
• Cosθ = Cos 30° = √3/2
• T = 4.9 s

$R = 49 \times \frac{ \sqrt{3} }{2} ×4.9$

$R= 120.05 \sqrt{3} \: m$

ANSWER :

a) 30.0125 m

b) 4.9 s

c) 120.05 √3 m