A student drops a rock from a bridge to the water 12m below. How many seconds does it take the rock to hit the water?
1 answer:
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2.57 joule energy lose in the bounce .
<u>Explanation</u>:
when ball is the height of 1.37 m from the ground it has some gravitational potential energy with respect to hits the ground
Formula for gravitational potential energy given by
Potential Energy = mgh
Where ,
m = mass
g = acceleration due to gravity
h = height
Potential energy when ball hits the ground
m= 0.375 kg
h = 1.37 m
g = 9.8 m/s²
Potential Energy = 5.03 joule
Potential energy when ball bounces up again
h= 0.67 m
Potential Energy = 2.46 joule
Energy loss = 5.03 - 2.46 = 2.57 joule
2.57 joule energy lose in the bounce
Answer:
3.62m/s and 2.83m/s
Explanation:
Apply conservation of momentum
For vertical component,
Pfy = Piy
m* Vof (sin38) - m*Vgf (sin52) = 0
Divide through by m
Vof(sin38) - Vgf(sin52) = 0
Vof(sin38) = Vgf(sin52)
Vof (sin38/sin52) = Vgf
0.7813Vof = Vgf
For horizontal component
Pxf= Pxi
m* Vof (cos38) - m*Vgf (cos52) = m*4.6
Divide through by m
Vof(cos38) + Vgf(cos52) = 4.6
Recall that
0.7813Vof = Vgf
Vof(cos38) + 0.7813 Vof(cos52) = 4.6
0.7880Vof + 0.4810Vof = 4.
1.269Vof = 4.6
Vof = 4.6/1.269
Vof = 3.62m/s
Recall that
0.7813Vof = Vgf
Vgf = 0.7813 * 3.62
Vgf = 2.83m/s
