Question

A ball is thrown from the top of a tower of height 63.8 m with a velocity of $30m{s}^{-1}$ at an angle of ${37}^o$ above the horizontal. The angle formed by the velocity of the stone with the horizontal when it hits the ground is (take g = $10m/s^2$)

• A. $ta{n}^{-1}\left(\frac{5}{3}\right)$
• B. $ta{n}^{-1}\left(\frac{3}{4}\right)$
• C. $ta{n}^{-1}\left(\frac{4}{3}\right)$
• D. $ta{n}^{-1}\left(\frac{3}{2}\right)$

Answer 1

The correct option is A $ta{n}^{-1}\left(\frac{5}{3}\right)$

A ball is thrown from the top of a tower of height $63.8m$ with a velocity of $30m/s$ at an angle of ${37}^o$ above the horizontal.

Given : $g=10m/s^2,h=63.8m,u=30m{s}^{-1},\theta ={37}^o$

Formula: for vertical motion

$h=u\sin \theta t-\frac{1}{2}g{t}^2$

Sol: Time taken to reach the ground,

$h=u\sin \theta t-\frac{1}{2}g{t}^2$

$\Rightarrow 63.8=\left(30\sin {37}^o\right)t-v_2\times 10\times t^2$

$\Rightarrow 63.8=18t-5t^2$

$\Rightarrow 5t^2-18t-5t^2$

$\Rightarrow t=5.8sec$

X-component:

Initial velocity, $u=u\cos theta$

$=u\cos {37}^o$

$=30\times 0.80=24m/s$

Velocity ofter $t\sec ,v=u+at$

$=2u+(0\times 5.8)$ ($\because$ acceleration along horizontal direction)

$=24m/s$

Y-component

Initial velcoity, $u^{\prime }=u\sin \theta =u\sin {37}^o$

$=30\times 0.60=18m/s$

velocity after $t\sec ,v^{\prime }=u^{\prime }+a^{\prime }t$

$=18+(-10)\left(5.8\right)$

$=-40m/s$

So, velocity after $t\sec$ in vector form,

$v=24\hat{i}-40\hat{j}$

Angle formed by the velocity of the stone with the horizontal

${\theta }^{\prime }=ta{n}^{-1}\left(\frac{V_y}{V_x}\right)=ta{n}^{-1}\left(\frac{40}{24}\right)$

${\theta }^{\prime }=ta{n}^{-1}\left(\frac{5}{3}\right)$

Option A