Question

A baseball is thrown from a rooftop with an initial downward velocity of magnitude v0= 5.38 m/s. The rooftop is at height h= 17.2 meters above the ground. Use a coordinate system where upwards is positive.

Part (a) Find the vertical component of the final velocity, vf,y, in meters per second, with which the ball hits the ground.
vf,y = - 19.1m/s
Part (b) If we wanted the baseball's final speed to be exactly 27.3 m/s, from what height, h_new, in meters, would we need to throw it with the same initial velocity?
h_new = 36.60m
***Part (c) If the height is fixed at 17.2 m, but we wanted the baseball's final speed to be 32.7 m/s, what would the vertical component, in meters per second, of the initial velocity, vi,y, need to be?

Answer 1

Answer: -27.06 m/s.

Step-by-step explanation:

Part (a)

Given:

v0 = -5.38 m/s (negative because it's downward)

h = 17.2 m

g = 9.81 m/s^2 (acceleration due to gravity)

Using the kinematic equation:

vf^2 = v0^2 + 2gh

vf = sqrt(v0^2 + 2gh)

Plugging in the values:

vf = sqrt((-5.38)^2 + 2(9.81)(17.2))

vf = sqrt(28.9444 + 336.912)

vf = sqrt(365.8564)

vf = 19.1 m/s

Since the direction is downward, the velocity is negative:

vf,y = -19.1 m/s

Part (b)

Given:

vf = 27.3 m/s

v0 = -5.38 m/s

Using the kinematic equation:

vf^2 = v0^2 + 2gh_new

Rearranging for h_new:

h_new = (vf^2 - v0^2) / (2g)

Plugging in the values:

h_new = (27.3^2 - (-5.38)^2) / (2(9.81))

h_new = (746.29 - 28.9444) / 19.62

h_new = 717.3456 / 19.62

h_new = 36.60 m

Part (c)

Given:

h = 17.2 m

vf = 32.7 m/s

Using the kinematic equation:

vf^2 = vi^2 + 2gh

Rearranging for vi:

vi^2 = vf^2 - 2gh

vi = sqrt(vf^2 - 2gh)

Plugging in the values:

vi = sqrt(32.7^2 - 2(9.81)(17.2))

vi = sqrt(1068.89 - 336.912)

vi = sqrt(731.978)

vi = 27.06 m/s

Since the direction is downward, the velocity is negative:

vi,y = -27.06 m/s

So, the vertical component of the initial velocity, vi,y, would need to be -27.06 m/s.