A 800 kg projectile is fired vertically upward with an intitial velocity of 50 m/s. Assuming that no friction is present, how high above the ground will the projectile travel?
Well, at the apex, the velocity is zero, so
Vf^2=V0^2 -2g*h
solve for h.
To find the height the projectile will reach, we can use the equations of motion from classical mechanics. The key equation we need is:
v^2 = u^2 + 2as
where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
s is the displacement.
Given:
u = 50 m/s (initial velocity),
v = 0 m/s (final velocity at the highest point, when the projectile momentarily stops),
a = -9.8 m/s^2 (acceleration due to gravity, assuming upward is positive),
s = ? (displacement, which represents the height we are trying to find).
Plugging in the values into the equation, we have:
0^2 = 50^2 + 2*(-9.8)*s
Simplifying the equation:
0 = 2500 - 19.6s
Rearranging the equation to solve for s:
19.6s = 2500
s = 2500 / 19.6
s ≈ 127.55 meters
Therefore, the projectile will reach approximately 127.55 meters above the ground.