Question

A projectile is fired horizontally from a gun that is 38.0 m above flat ground, emerging from the gun with a speed of 300 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

Answer:

a)t=2.78 sec

b)R=835.03 m

c)$v_y=27.27\ m/s$

Explanation:

Given that

h= 38 m

u=300 m/s

$y=y_o+u_yt-\dfrac{1}{2}gt^2$

here given that

$u_y=0$

$y_o=38$

The finally y=0

So

$y=y_o+u_yt-\dfrac{1}{2}gt^2$

$0=38+0-\dfrac{1}{2}9.81\times t^2$

t=2.78 sec

Horizontal distance,R

R= u x t

R=300 x 2.78

R=835.03 m

The vertical component of velocity before strike

$v_y=gt$

$v_y=9.81\times 2.78$

$v_y=27.27\ m/s$