Question

A plane.diving with a constant speed at an angle of 53.0 with the vertical,releases a projectile at an altitude of 730m.The projectile hits the ground 5.00s after release.(A) WHat is the speed of the plane?

Answer 1

The velocity of the plane at the instant that the projectile leaves is 202.5m/s

Explanation:

We consider that the projectile's vertical motion as a body falling under gravity with initial velocity.
Assumption 1: the projectile leaves tangential to the plane
Assumption 2: at the instant the projectile leaves the plane, the relative velocity between them is 0
Here S = 730m
t = 5s
angle = 53degrees to the vertical so it is 90 - 53 = 37 degrees to the horizontal
taking acceleration due to gravity as 9.8m/s^2
the vertical component of the initial velocity is usin37 = 0.6*u
now we use `S = u*t + (g*t*t)/2`
we have `730 = 0.6*u*5 + (9.8*5*5)/2`
that is `730 = 3*u + 122.5`
rearranging we get `u = (730 - 122.5)/3`
so `u = 202.5m/s`