A 8.57 kg ball is thrown upward with an initial speed of 14.85 m/s. Ignore air r
ID: 1433265 • Letter: A
Question
A 8.57 kg ball is thrown upward with an initial speed of 14.85 m/s. Ignore air resistance when answering the following questions. What is the ball's initial kinetic energy? Tries 0/6 What is the ball's kinetic energy when it reaches its maximum height? Tries 0/6 What is the ball's gravitational potential energy (relative to Earth's surface) when it reaches its maximum height? Tries 0/6 What is the ball's maximum height? Tries 0/6 What is the ball's kinetic energy just before it returns to your hand? Tries 0/6Explanation / Answer
fo rthe first bit , given m = 8.57kg , speed i.e. V = 14.85m/s
- KE of the ball,
We know that K.E = 1/2 mv^2
= 1/2 (8.57)(14.85)^2
=944.93 J
therefore Initial KE of the ball is =0.95KJ
The KE of the ball at its maximum height is s = Ut + (at^2)/2
to Find t lets use V = U + at
where
V ( the final velocity at the highest point)= 0 m/s ( the spped of the object reduces to zero when it reaches the highest point)
u ( the initial velocity) = 14.85m/s
a= - 9.8 m/s^2 ( negative sign since the boby id moving against gravity)
substituting these valuse in the above equation we get ,
0 = 25.2 - 9.8 * t
t = ( 0 - 14.85) / ( - 9.8)
= 1.51seconds
now to find how high it rises ,use the equation
S = Ut + (a* t * t )/2
= 14.85 * 1.51+ ( - 9.8 * 1.51* 1.51 ) / 2
= 12.76 meters
time taken to reach the highest point = 1.51 seconds
time taken for the ball to hit the ground after it reaches its highest point = 1.51seconds
Potential Energy of the ball at its maximum height is Ep = mgh
= (8.57 * 9.8 * 12.76 )
= 1071.66136 KJ
Therefore, Potential Energy of the ball at its maximum height is = 1.071 J
Ball's KE just before it reaches your hand is : V^2 - U^2 = 2gS
V^2 - (14.85)^2 = 2*9.8*12.76
V^2 = 2*9.8*12.76*(14.85)^2
V = 234.84J
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