1. A ball rolls off a table with a horizontal velocity of 8 m/s. If it takes 0.3

Question

1.A ball rolls off a table with a horizontal velocity of 8 m/s. If it takes 0.3 s to reach the floor:

(a) What is the vertical component of the ball's velocity just before it hits the floor?
(b) What is the horizontal component of the ball's velocity just before it hits the floor?
Use g = 10 m/s2

2.A ball rolls off a platform that is 6.8 meters above the ground. The ball's horizontal velocity as it leaves the platform is 7.4 m/s.

(a) How much time does it take for the ball to hit the ground?
(b) How far from the base of the platform does the ball hit the ground?
Use g = 10 m/s2

3.A projectile is fired at an angle such that the vertical component of its velocity and the horizontal component of its velocity are both equal to 16 m/s.

(a) How long does it take for the ball to reach its high point?
(b) What is the vertical distance at this time?
(c) What is the horizontal distance at this time?
Use g = 10 m/s2

Explanation / Answer

1)

a) when the ball falls it took 0.3 sec and let acceleration due to gravity = 10

velocity before it hits ground = 0.3*10 = 3 m/s(v=a*t)

b) horizontal velocity does not change as gravity is the only force and it acts downwards

velocity = 8 m/s

2)

a)height =6.8

downward velocity initialy = 0 so using distance = (a*t^2)/2 a=g=10

6.8 = 5*t^2 = > t=squareroot of(1.36)

t=1.166 sec

b)distance horizontal = v*t = 7.4*1.166 = 8.6284

3)

a)vertical component = 16 m/s

a=-g = -10 as body is moving upward = >

time for velocity = 0 = > 0-16 = -10 *t = > t=1.6s

b)vertical distance s = v*t + (a*t^2)/2 v=16 , a=-g = -10 , t=1.6

s = 12.8 m

c)horizontal distance = v*t = 16*1.6 = 25.6 m

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