(5%) Problem 20: A ball is thrown from a rooftop with an initial downward veloci

Question

(5%) Problem 20: A ball is thrown from a rooftop with an initial downward velocity of magnitude vo- 0.65 m/s. The rooftop is a distance above the ground, h - 16 m. In this problem use a coordinate system in which upwards is positive Otheexpertta.com 33% Part (a) Find the vertical component of the velocity, vry, in meters per second, with which the ball hits the ground Grade Summary Deductions Potential 0% 100% sin0 cosO cotan asin0 tan() acosO atanOacotanO sinh0 coshO tanh cotanh Submissions Attempts remaining: 200 (0% per attempt) detailed view | | 0 END Degrees Radians CE DEL CLEAR Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 4 Feedback: 0% deduction per feedback. 33% Part (b) If we wanted the ball's final speed to be exactly 27.3 m/s, from what heght hne (in meters) would we need to throw it with the same initial velocity? 33% Part (c) If the height is fixed at 16 m, but we wanted the ball's final speed to be 32.7 m/s, what would the vertical component of the initial velocity viy need to be, in meters per second?

Explanation / Answer

Given that

V0 = initial velocity = -0.65 m/sec (negative since velocity is downward)

a = -g = -9.81 m/sec^2

h = height of building = -16 m

Now Using equation

Vfy^2 = V0^2 + 2*a*h

Vfy = sqrt ((-0.65)^2 + 2*(-9.81)*(-16))

Vfy = -17.73 m/sec (Again negative Since velocity direction is downward)

Part B.

If Vfy_new = 27.3 m/sec

Using same equation for h_new

Vfy_new^2 = V0^2 + 2*a*h_new

h_new = [(-27.3)^2 - (-0.65)^2]/(2*(-9.81))

h_new = -37.96 m

Height of building = 37.96 m = 38.0 m

Part C.

Now when Vfy = -32.7 m/sec

h = -16 m then

Viy = ?

Vfy^2 = Viy^2 + 2*a*h

Viy^2 = Vfy^2 - 2*a*h

Viy = sqrt [(-32.7)^2 - 2*(-9.81)*(-16)]

Viy = -27.48 m/sec = -27.5 m/sec

See that in each part answer will be negative, since given that upward is positive

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