Question

A projectile is fired horizontally from a gun that is 70.0 m above flat ground, emerging from the gun with a speed of 250 m/s.
(a) How long does the projectile remain in the air?
s

(b) At what horizontal distance from the firing point does it strike the ground?
m

(c) What is the magnitude of the vertical component of its velocity as it strikes the ground?

Answer 1

Start by drawing a labeled diagram

Answer 2

#a

Since the projectile is thrown horizontally, vertical component of the velocity will be zero. Thus you can calculate the time of flight (t) by using

\(\sf s=ut+\frac {1}{2}at^2\)
Take,
s=70m
u=0
a=9.8

Answer 3

Once you find t in the above question, use it to find horizontal distance (s) by using the formula

\(\sf s=ut+\frac{1}{2}at^2\)
take,
u=250m/s
t= as calculated in the above problem
a= 0 (because there is no acceleration in the horizontal direction)

Answer 4

#c

Finally calculate the final velocity (v) by using the equation v=u+at
Take,
u=0
t= as calculated in the first problem
a=9.8

Answer 5

thanks that was very helpful for many questions not just this one