vertical:
hf=hi+vi*sin53*20-4.9*20^2
hf=0 hi=800
solve for vi
Horizontal:
horiz distance=vi*cos53*20
max height:
mgh=1/2 m (vi*sin53)^2
solve for h, then add initial 800m altitude.
Velocity at impact
1/2 m vi^2+mg*800=1/2 m vf^2
solve for vf
your teacher is too easy.
A bomber, flying upwards at an angle 53 with verticql releases a bomb at 800m altitude. The bomb strikes the ground 20s after release. Find
A--velocity of bomber at time of release of bomb
B--maximum height attained by bomb
C--horizontal distance covered by bomb before reaching gorund
D--velocity of bomb when it strikes the ground
2 answers
Madharchod h Tu sala bhosdi
0 answers