(Assume g = 9.80 m/s 2 .) (a) A small block is released from rest at a height of
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Question
(Assume g = 9.80 m/s2.)
(a) A small block is released from rest at a height of 3.65 m above the ground and moves downward in free fall.
(i) How long does it take the block to reach the ground?
(ii) What is the speed of the block just before it strikes the ground?
(b) A frictionless incline is 5.00 m long (the distance from the top of the incline to the bottom, measured along the incline). The vertical distance from the top of the incline to the bottom is 3.65 m. A small block is released from rest at the top of the incline and slides down the incline.
(i) How long does it take the block to reach the ground?
(ii) What is the speed of the block just before it strikes the ground?
Explanation / Answer
A small block is released from rest at a height of 3.65 m
Therefore the vertical displacement of the block is (S) = 3.65 m
The block is released from rest so the initial velocity of the block (u) = 0 m/s
The body has an acceleration due to gravity (a) = 9.8 m/s^2
(ii) What is the speed of the block just before it strikes the ground?
v^2 = u^2 + 2as where v is the speed of the block just before it strikes the ground
v^2 = 0^2 + 2(9.8)(3.65)
v^2 = 71.54
v = 8.45 m/s
(i) How long does it take the block to reach the ground?
v= u + at where t is time
8.45=0 + 9.8t
t = 0.862 sec
b ) the block in the the vertical direction has an acceleration of 9.8m/s2
the block in the vertical direction has a displacement of 3.65 m
the block started drom rest so the initial velocity is 0 m/s
(i) s= ut +at^2/2
3.65 = 0t + 4.9t^2
t^2 = 0.74
t = 0.86 s
too big sum for the second part find vertical and horizontal velocity of the block square em add em under a root
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