Question

A ball is projected upwards from the top of a tower with a velocity of $50m{s}^{-1}$ making an angle of $30$ with the horizontal. The height of the tower is $70m$. After how much time from the instant of throwing will the ball reach the ground?

• A. $2s$
• B. $5s$
• C. $7s$
• D. $9s$

Answer 1

Answer: (C)

$7s$

time of flight $t=\frac{2vsin\theta }{g}=\frac{2\times 50\times sin30}{10}=5s$
When the ball reaches the tower, on its downward motion, its speed is $50\sin {30}^0=25m/s$

now the vertical component of the velocity is same at the start of the flight and end of the flight therefore $v_y=vsin\theta =50sin{30}^o=25m/s$

$S=ut+\frac{1}{2}g{t}^2=25t+\frac{1}{2}g{t}^2$

$70=25t+5t^2$

$t^2+5t-14$

$t^2+7t-2t-14$

$t=2,-7$ are the roots, $2$ is taken

hence total time taken is $T=5+2=7s$