Question

from the top of a tower of highest 490m a shell is fired horizontal with a velocity 100m/s. At what distance from the bottom of the tower, the shell will not the ground​

Answer 1

Assuming no drag :

H = 490m

USince the velocity in the vertical direction is zero, therefore :

U(vertical) =0m/s

Applying Kinematic Equation : H= ut+1/2at2

Plug the values of H(vertical height)=490m.

U(velocity in y direction)=0m/s

a=9.8m/s2

We will get t=10s

Which implies that the shell will take 10 seconds to reach the ground after it is fired.

Since there is zero acceleration in the horizontal direction therefore the velocity (horizontal) must remain unchanged, therefore velocity throughout the projectile motion will be 100m/s in the horizontal direction.

Therefore the distance from the bottom must be velocity(horizontal)∗time=

100∗10=1000m