Discussion Forum

1)

A ball is thrown at a speed of 20 m/s at an angle of $30^{0}$ with the horizontal. The maximum height reached by the ball is

[Use g=10 m/s²]

Answer:

Option D

Explanation:

 Given,

 speed of ball (u) =20 m/s

 Angle ($\theta$)= $30^{0}$

 We know that,

   $H= \frac{u^{2}\sin^{2}\theta}{2g}$

$H= \frac{(20)^{2}(\sin30)^{2}}{2\times10}$

 $H= \frac{20\times20\times\left(\frac{1}{2}\right)^{2}}{20}$

 $H= 20 \times \frac{1}{4}$

 H=5 m