Question

A body is thrown with velocity of $40m/s$ in a direction making an angle of ${30}^{\circ }$ with the horizontal. Calculate i) Horizontal range, ii) Maximum height and iii) Time taken to reach the maximum height.?

Answer 1

$u=40m/s$ and $\theta ={30}^0$

1. Horizontal range will be $R=\frac{u^{2}Sin2\theta }{g}=\frac{40\times 40\times \sqrt[2]{23}}{10\times 2}=80\sqrt[2]{23}m$

2. maximum height $H=\frac{u^{2}Si{n}^2\theta }{2g}=\frac{1600\times 1}{20\times 4}=20m$

3. half of the time of $flight$ i.e $\frac{T}{2}=\frac{uSin\theta }{g}=\frac{20\times 1}{10\times 2}=1second$