Question

An elevator is moving upward at a constant speed of 6.6 m/s. At 40 m above the level ground, a tennis ball is shot from the elevator at a speed of 49 m/s and an angle of 48° above horizontal, relative to the elevator. Rel. velocity of the ball to ground?

The question wants to know the horizontal and vertical velocities of the ball relative to the ground, and then how long it takes for the ball to make contact with the ground.

Answer 1

I figured it out! 9.63 seconds

Explanation:

We know the ball is fired at 48 degrees from the horizontal relative to the elevator. The elevator has a constant vertical velocity. We have to factor this into our calculation of the vertical velocity of the ball relative to the ground.

`49sin41+6.6 = 43.01 " m/s"`

Plugging that into `y_f=y_0+v_0t+1/2at^2`...

`0=40+43.01t+1/2(-9.8)t^2`

I found the value of `t` by plugging the equation into my calculator and finding the positive value where the line crosses the y-axis.

graph{(1/2*-9.8)x^2+43.01x+40 [-2.936, 12.204, -2.82, 4.75]}