A ball starts from rest and accelerates at 0.480 m/s^2 while moving down an incl

Question

A ball starts from rest and accelerates at 0.480 m/s^2 while moving down an inclined plane 9.15 m long. When it reaches the bottom, the ball rolls up another plane, where, after moving 14.40 m, it comes to rest. What is the speed of the ball at the bottom of the first plane? How long does it take to roll down the first plane? What is the acceleration along the second plane? What is the ball's speed 8.10 m along the second plane? A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically upwards from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground. (Use h and g as appropriate in your equation.)

Explanation / Answer

Multiple questions asked I am allowed to solve only 1 question at a time according to the chegg guidelines.

14 ) we need to find the speed of the ball

we know the equation

v^ 2 = u ^ 2 + 2as

v^2 = 0 + 2 * 0.480 m/s^2 * 9.15 m

v = 2.96378 m/s ================================part a )

part b) we need to find the tim taken

we know the eqn

v = u + at

t = 2.96378 - 0 / 0.480 = 6.1745445 seconds ==========part b)

Part c) now we need to find the acceleration

we will use the eqn

v^ 2 = u ^ 2 + 2as

0 = 2.96378 m/s ^ 2 + 2 * a * 14.40 m

a = -0.305 m/s^2 ===================part c)

Part d ) we need to find the ball speed

again we use same eqn

v^ 2 = u ^ 2 + 2as

v^2 = 2.96378 m/s ^ 2 + 2 *  -0.305 m/s^2 * 8.10 m

v = 1.960355 m/s ==================part d)

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