Question

A cricket ball is thrown from the earth at a speed of 50 m/s in a direction, making an angle of 30°
with the horizontal. Calculate:
(a) Maximum height
(b) Time of flight of the ball to return to the earth (g = 9.8 m/s²).​

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\large\red{\boxed{\sf H_{max}=31.88m }}$

$\large\red{\boxed{\sf T=5.1sec }}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Speed (v) = 50 m/s

• Angle $\sf{\theta}$=30°

$\large\underline\pink{\sf To\:Find: }$

• Maximum Height $\large{\sf H_{max}=?}$

• Time of flight = ?

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a) Maximum Height :

We know that ,

$\large{♡}$$\large{\boxed{\sf H_{max}={\frac{{u}^{2}{sin}^{2}\theta}{2g}}}}$

On Putting value :

$\large\implies{\sf {\frac{{50}^{2}×{sin}^{2}30°}{2×9.8}}}$

Sin 30° = 1/2

$\large\implies{\sf {\frac{2500×1}{4×19.6}}}$

$\large\implies{\sf {\frac{2500}{78.4}}}$

$\large\implies{\sf 31.88m}$

$\huge\red{♡}$$\large\red{\boxed{\sf H_{max}=31.88m }}$

_________________________________________

b) Time of flight of the ball to return to the earth .

We know that ,

$\large{♡}$$\large{\boxed{\sf T={\frac{2usin\theta}{g}}}}$

$\large\implies{\sf {\frac{2×50×sin30°}{9.8}}}$

$\large\implies{\sf {\frac{2×50×1}{2×9.8}}}$

$\huge\red{♡}$$\large\red{\boxed{\sf T=5.1sec }}$

Answer 2

The maximum height reached by the ball is 31.88 m and its time of flight is 5.1 s.

Explanation:

Given that,

Initial speed of the cricket ball, u = 50 m/s

Angle of projection with the horizontal, $\theta=30^{\circ}$

(a) The maximum height reached by the ball is given by formula as follows :

$h=\dfrac{u^2\sin^2\theta}{2g}\\\\h=\dfrac{(50)^2\sin^2(30)}{2\times 9.8}\\\\h=31.88\ m$

So, the maximum height reached by the ball is 31.88 meters.

(b) Time of flight is given by :

$t=\dfrac{2u\sin\theta}{g}$

$t=\dfrac{2\times 50\sin(30)}{9.8}\\\\t=5.1\ s$

So, the maximum height reached by the ball is 31.88 m and its time of flight is 5.1 s.

Learn more,

Projectile motion

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