THIS IS THE COMPLETE QUESTION BELOW
With what minimum speed must you toss a 130 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release the ball 1.1 m above the ground
And what speed does the ball hit the ground? Solve this problem using energy.
Answer
a)minimum speed must you toss a 130 g is 15.9090m/s
b)speed the ball hit the ground is 16.57m/s
Explanation:
a)We know that For any closed/isolated system, the total energy is CONSERVED.
K.E. lost by the ball=The change in P.E of the ball at 1.1m above the ground as well as the P.E. of the ball at 14 m-high roof
This statement can be expressed as the expression below from K.E and P.E energy formula
P.E. = mgh
K.E. = (1/2)mv^2
Therefore,
(mgh1 - mgh2)=(1/2)mv^2
Where h2=the ball height above the ground=1.1m
h1=ball height at roof of the gymnasium= 14m
Then if we substitute we have
[(10) x (0.14) x (9.81)] - [(1.1) x (0.14) x (9.81)] = (1/2)(0.14)(v^2)
16.45137=0.065V^2
V=15.9090m/s
minimum speed must you toss a 130 g is 15.9090m/s
b)To calculate the speed the ball hit the ground?
This is the highest point (14m-high roof),and the type of the energy the ball possesses is Po.tential energy only.
At the lowest point (ground), the energy the ball possesses is K.E. only.
P.E at 10m-high roof = K.E. at ground.
(14) x (0.13) x (9.81) = (1/2) x (0.13) x v^2
17.8542= 0.065V^2
V= 16.57
Therefore,And speed the ball hit the ground is 16.57m/s
