A ball is thrown from the top of a tower with an initial velocity of 10 m//s at an angle of `30^(@)`above the horizontal. It hits the ground at a distance of 17.3 m from the base of the tower. The height of the tower `(g=10m//s^(2))` will be

Options

A. 5 m
B. 20 m
C. 15 m
D. 10 m

Answer

D

Explanation

Here, `theta=30^(@),u=10ms^(-1),R=17.3m,g=10ms^(-2)` For horizontal motion, `R=ucosthetat`
or `t=(R)/(ucostheta)=(17.3)/(10cos30^(@))=(17.3xx2)/(10xxsqrt3)=(17.3xx2)/(10xx1.73)=2s`
For vertical motion, `y=usinthetat-(1)/(2)g""t^(2)`
`y=10sin30^(@)xxs-(1)/(2)xx10xx2^(2)=10-20=-10m`
Height of tower =10m.