Question

A cricketer can throw a ball to a maximum horizontal distance of 100 m . find maximum velocity with which cricketer can throw the ball​

Answer 1

Answer:

☆Solution:

》horizontal distance during projectile motion is called as range(R).

》We know that,

R = $\frac { u^{2} sin(2\theta)}{g}$

where u = initial velocity by which the ball thrown.

g = gravity (10m/s2)

》range will be maximum when$\bold{\theta}$ will be 45°. Bcz sin(2*45°)= sin(90°)

and sin90 is maximum = 1

》so formula for maximum range :

R = $\frac {u^{2}}{g}$

• R = $\frac {u^{2}}{g}$

• 100 = $\frac {u^{2}}{10}$

• u =$\bold{10 \sqrt{10}}$

So,the maximum velocity by which the cricketer throw the ball is $\bold{10\sqrt{10}}$.

$\bold{♧hope\;it\;helps\;!\;♧}$