Question

A ball is thrown from the top of a tower with as initial velocity of 10 $m{s}^{-1}$ at an angle of ${30}^0$ with the horizontal. If it hits the ground at a distance of 17.3 m from the base of the tower, then the height of the tower will be (Given g = 10 $m{s}^{-2}$)

• A. 30 m
• B. 20 m
• C. 43.75 m
  
  
• D. 15 m

Answer 1

The correct option is C 43.75 m



$H=\frac{u^2{\sin }^2\theta }{2g}=\frac{100}{20}\times \frac{1}{4}=1.25m$

$R=\frac{u^2\sin 2\theta }{g}=\frac{100}{10}\times \frac{\sqrt{3}}{2}=5\sqrt{3}m$

$t_{AC}=T=\frac{2u\sin \theta }{g}=\frac{2\times 10}{10}\times \frac{1}{2}=1s$

$t_{BC}=0.5s$

Vertical velocity at B,$v_{By}=0m/s$

Horizontal velocity at B,$v_{Bx}=u\cos {30}^{\circ }=5\sqrt{3}$

$R+x=17.3m$

$x=17.3-5\sqrt{3}=8.64m$

$y=\frac{R}{2}+x=12.97m$

$t_{BD}=\frac{y}{V_{Bx}}=\frac{12.97}{5\sqrt{3}}=1.5s$

$t_{CD}=t_{BD}-t_{BC}$

$=1.5-0.5=1s$

$V_{Dy}=V_{By}+2a{t}_{BD}=0+20\times 1.5$

$V_{Dy}=30m/s$

$V_{Dy}^2-V_{By}^2=2as$

$900=20(H+h)=20H+20h$

$20h=900-20H$

$900-25=875$

$h=\frac{875}{20}=43.75m$