Physics Question (ASAP)

Question

Consider a dodgeball thrown straight up in the air from an initial height of 0.550 m above the ground with an initial velocity of 16.6 m/s. Assume the ball is in freefall.

A. Predict the velocity of the soccer ball after 1.10 s of free fall.

B. Predict the highest point above the ground the soccer ball reaches.

C. Is the dodgeball going faster, slower or the same when it hits the ground compared to it’s initial velocity?

D. Predict the velocity of the soccer ball at the apex of its freefall trajectory.

Answer 1

h(t) = -4.9t^2 + vt + ho = -4,9t^2 + 16.8t + 0.550

where h= height in meters at time t

v = initial velocity, m/sec

ho = initial height

t = time in seconds

velocity after 1.1 seconds is

h(1.1) = -4.9(1.1)^2 + 16.8(1.1) + 0.55

highest point is found by taking the derivative h'(t) and setting it equal to zero, solved for t, then plug that t value into the h(t) equation to get the height.

h' = -9.8t + 16.8 = 0. t=16.8/9.8 = 8.4/4.9 =

max h = -4.9(8.4/4.9)^2 +16.8(8.4/4.9) + 0.55 meters

faster when it hits the ground. Had it started at ground level then the velocities would have been the same, but gravity added some additional velocity as it fell below 0.55 meters

velocity at the apex? at max height? that's zero,