A ball is thrown horizontally from the top of a 65 m building and lands 115 m from the base of the building. Ignore air resistance, and use a coordinate system whose origin is at the top of the building, with positive y upwards and positive x in the direction of the throw.

(a) How long is the ball in the air in seconds?

(b) What must have been the initial horizontal component of the velocity in m/s?

(c) What is the vertical component of the velocity just before the ball hits the ground in m/s?

(d) What is the magnitude of the velocity of the ball just before it hits the ground in m/s?

Question

Mekhi

Physics

18 July, 05:35

A ball is thrown horizontally from the top of a 65 m building and lands 115 m from the base of the building. Ignore air resistance, and use a coordinate system whose origin is at the top of the building, with positive y upwards and positive x in the direction of the throw.

(a) How long is the ball in the air in seconds?

(b) What must have been the initial horizontal component of the velocity in m/s?

(c) What is the vertical component of the velocity just before the ball hits the ground in m/s?

(d) What is the magnitude of the velocity of the ball just before it hits the ground in m/s?

+3

Answers (1)

Know the Answer?

Chase Kim · Answer 1

The answer to your question is:

a) t = 3.64 s

b) vox = 31.59 m/s

c) vy = 35.71 m/s

d) v = 47.67 m/s

Explanation:

a) To calculate the time, we know that voy = 0 m/s so we this this formula

h = voy + 1/2 (gt²)

65 = 0 + 1/2 (9.81) (t²)

65 = 4.905t²

t² = 65/4.905 = 13.25

t = 3.64 s

b) To calculate vox we use this formula vox = d/t; vox is constant

vox = 115/3.64 = 31.59 m/s

c) To calculate voy we use the formula vy = voy + gt

but voy = 0

vy = gt = 9.81 x 3.64 = 35.71 m/s

d) To calculate v we use the pythagorean theorem

c2 = a2 + b2

c2 = 31.59² + 35.71² = 997.92 + 1275.20

c2 = 2273.12

c = 47.67 m/s