Question

T) 6073
A bomber flying upward at an angle of
53° with the vertical releases a bomb at
an altitude of 800 m. The bomb strikes
the ground 20 s after its release. If
g=10ms-2, the velocity at the time of
release of the bomb in ms-1 is
1) 400 2) 800 3) 100 4) 200​

Answer 1

Explanation:

u=?

s=800 m

t=20 sec

g=10 m/ sec^2

$s = ut + \frac{1}{2} g {t}^{2} \\ \\ 800 = u(20) + \frac{1}{2} (10)( {20)}^{2} \\ \\ 800 = 20u + \frac{1}{2} (4000) \\ \\ 800 = 20u + 2000 \\ \\ 20u = 1200 \\ \\ u = 60 \: \frac{m}{sec}$