A baseball is thrown from the roof of 23.5 m -tall building with an initial velo
ID: 1469400 • Letter: A
Question
A baseball is thrown from the roof of 23.5 m -tall building with an initial velocity of magnitude 12.3 m/s and directed at an angle of 54.8 above the horizontal.
A)What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance
B)What is the answer for part (A) if the initial velocity is at an angle of 54.8 below the horizontal?
C)If the effects of air resistance are included, will part (A) or (B) give the higher speed?
The part (A) will give the higher speed. or The part (B) will give the higher speedExplanation / Answer
As the baseball is thrown from the roof of a building of height 23.5 m with an initial velocity of magnitude 12.3 m/s , the base ball possesses kinetic energy (KE) as well as potential energy (PE)
Initial KE=(1/2)mu^2 =(1/2)m*12.3*12.3=75.645m J
Initial PT =mgh=m*9.8*23.5= 230.3 m J
Initial total energy E= KE + PE
Initial total energy E= 75.645 m + 230.3 m =305.945m J
Initial total energy E= 305.945m J
When the ball strikes the ground, its final PE is zero and its final KE is (1/2)mV^2, where V is its velocity when it strikes the ground.
Final total energy E! = final KE + final PE
Final total energy E! =(1/2)mV^2 + zero
Ignoring air resistance, the total energy remains conserved
Final total energy E! = initial total energy E
(1/2)mV2= 305.945m J
cancelling 'm'
V2 =2*305.945 =611.89
V = 24.7364 m/s
Therefore, the speed of the ball just before it strikes the ground is 23.67 m/s.
B)
I didn't use the launching angle to solve part a), which means that it doesn't affect the result, so
v = 24.7364 m/s
C)
In (b), the ball has a shorter path through the air, so it will be slowed down less than the ball in part (a), so (b) has the higher speed.
Hope this helps. :)
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