Question

You toss a ball from a balcony. When the ball leaves your hand, it is 10.0...

You toss a ball from a balcony. When the ball leaves your hand, it is 10.0 meters above the ground and has a velocity of (10.0 m/s, 45.0 degrees above the horizontal). Ignore air resistance. (let's take the x-direction to be aligned with the horizontal motion and the y-axis to be upward.)

A) How high above the ground does the ball go, and how much time does it take to reach max height?

B) Determine the total time of flight.

C) How far does the ball land from the base of the balcony?

D) What is the ball's impact velocity?

E) Find the direction of the velocity vector at the point of landing.

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Answer #1

a)

height above the ground

H = 10 + (10 sin 45)^2 / (2* 9.8)

H = 12.55 m

time taken to reach the max height

t = 10 sin 45 / 9.8 = 0.7215 s

======

b)

using 2nd equation of motion

0 = 10 + 10 sin 45 * t - 0.5 * 9.8* t^2

solving for t

t = 2.3218 s

=====

c)

x = 10 cos 45 * 2.3218 = 16.417 m

=====

d)

v^2 = 10^2 + 2 * 9.8* 10

v = 17.2 m/s

=====

e)

direction,

cos x = Vx / V

x = arccos ( 10 cos 45 / 17.2)

x = 65.726

======

Comment before rate in case any doubt, will reply for sure.. goodluck

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