a ball is rolled with a speed of 11.3 m/s along a horizontal section of a ramp t

Question

a ball is rolled with a speed of 11.3 m/s along a horizontal section of a ramp that is slanted upward. the end of the track is 0.345m above the ground. the ball leaves the end of the ramp at an angle of 52.5 degrees above the horizontal, following the path of projectile motion. ignore friction and air resistance.

a) find the maximun height to which the ball rises above the end of the track

b) how long after leaving the ramp does it take the ball to reach this maximum height?

c) what is the horizontal distance from the end of the ramp to a point directly beneath the maximum height of the ball?

please show detailed step by step answer including equations used, thanks!

Explanation / Answer

2)V at the end of ramp

again energy conservation

1/2 mv1^2 =mgh+1/2mV2^2

here h=0.345m

v2 on solving =10.99m/sec

so velocity which it leaves ramp=10.99m/s

horizontal component =10.99 cos 52.5 =6.69 m/se (responsible for horizontal distance)

vertical component=10.99 sin 52.5=8.71 m/sec(responsible for taking to max height)

so hmax=0.345 +(8.71^2/(2*9.81) )

=4,211 m<---ans

for time use

1st law of motion

=>0=8.71-9.8at

=>t=0.887 sec<---ans

horizontal distance=6.69*0.887=5.93m <---ans

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