Question

A projectile is fired horizontally from a gun that is 54.0 m above flat ground, emerging from the gun with a speed of 330 m/s. (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground?

Answer 1

Explanation:

Given

height of building (h)=54 m

Projectile velocity=330 m/s

initial vertical velocity$(u_y)=0$

Initial horizontal velocity$(u_x)=330 m/s$

Time taken to cover vertical height of 54 m

$h=ut+\frac{gt^2}{2}$

$54=0+\frac{9.81\cdot t^2}{2}$

$t=\sqrt{\frac{54\times 2}{9.81}}$

t=3.31 s

Horizontal distance traveled in this time

$R_x=u_x\times t$

$R_x=330\times 3.31=1094.94 m$

Vertical component of velocity when it hits the ground

v=u+at

$v=0+9.81\times 3.31=32.47 m/s$