Question

A body when projected at an angle ${60}^o$ with the horizontal covers a horizontal distance of $90$m in $5$ s. Instead, if the body is thrown upward with the same velocity. The maximum height attained by it is _____ m. $(Takeg=10{ms}^{-2})$

• A. $68.8$
• B. $64.8$
• C. $34.8$
• D. $36.4$

Answer 1

The correct option is B $64.8$

Let the velocity of projectile be $u$.

Angle of projection with the horizontal $\theta ={60}^o$

So, initial velocity in horizontal direction $u_x=u\cos \theta =u\times \cos {60}^o=0.5u$

Horizontal distance covered in 5 seconds is $90m$

Thus $0.5u=\frac{90}{5}$

$⟹u=36m/s$

Now the body is projected vertically upwards with a speed $u=36m/s$

Maximum height reached $H=\frac{u^2}{2g}=\frac{36\times 36}{2\times 10}=64.8m$