Question

A particle is projected from ground with some initial velocity making an angle ${45}^o$ with the horizontal. It reaches at height of 7.5 m above the ground while it travels a horizontal distance of 10 cm from the point of projection. The initial speed of the projection is

• A. 5 m/s
• B. 10 m/s
• C. 20 m/s
• D. 40 m/s

Answer 1

The correct option is C 20 m/s

Here in this case,

$\theta ={45}^o,g=10m/s^2$, Horizontal distance $\left(x\right)=10m$,

Vertical distance $\left(y\right)=7.5m$

$y=\left(\tan \theta \right)x$ $-\left(\frac{g}{2u^2{\cos }^2\theta }\right)x^2$

$\Rightarrow 7.5=(1\times 10)-\left(\frac{10}{2u^2\times \frac{1}{2}}\right)\times 100$

$\Rightarrow \frac{1000}{u^2}=2.5$

$\Rightarrow u^2=\frac{1000}{2.5}=400$

$\Rightarrow u=\sqrt{400}=20$ m/s.