Q. An aero-plane is flying with a constant speed `u` at an angle `theta` above the horizontal.A ball is dropped (with zero initial velocity w.r.t the plane) from the aero-plane.It hits the ground after time `t.` What is the height of the aero-plane above the ground at this instant?
A) `1/2 g(t^2)`
B) ` ` `usintheta t - 1/2g(t^2)`
C) `usintheta t+1/2g(t^2)`
D) `1/2g(t^2)-usintheta t`
The ball is released from the airplane at point A). It hits the ground at point B) after time t. See attached figure.
The vertical component Vy of the airplane speed in any point of the trajectory is
`Vy = u*sin(theta)`
At point A on the y axis the ball has the speed Vy (relative to the ground) and it falling under the gravitational acceleration g.
The distance H1 (from the release point A to the ground) is
`H1 = -Vyt + g*t^2/2 = -u*tsin(theta) + g*t^2/2`
The negative sign in the equation comes from the fact that the initial vertical speed Vy of the ball and g are opposite one to each other.
From the point A to point B (where the ball hits the ground) the plane vertical displacement is H2 (see the figure). The motion of the plane is uniform (no acceleration) on both axes x and y, therefore
`H2 = +u*t*sin(theta)`
Now the total vertical distance H (from the plane, to the ball when it hits the ground) is
`H =H1+H2 = g*t^2/2`
The correct answer is A)