Asked by Niki
An outfielder throws a baseball to home plate with a velocity of +15 m/s and an angle of 30°. When will the ball reach its highest altitude?
(Points : 3)
1.2 s
1.5 s
0.77 s
0.50 s
Answers
Answered by
Damon
Vi = 15 sin 30 = 7.5
v = Vi - gt
v = 0 at top
so
g t = 7.5
t = 7.5/9.8 = about .77 s
v = Vi - gt
v = 0 at top
so
g t = 7.5
t = 7.5/9.8 = about .77 s
Answered by
Devron
The vertical component of the balls velocity is the following:
15m/s*Sin 30=7.5m/s
The deceleration due to gravity is 9.8m/s. So, after 1s the ball will be falling back down to the ground, which eliminates the first two answer choices. Does it take a half a second for the ball's velocity to reach 0m/s, or does it take slightly more time for the ball's velocity to reach 0m/s? In half a second, the ball will decelerate by 4.8m/s, which means that it will still have a velocity of 2.7m/s. The best answer choice is 0.77s.
****You could have just used an equation, but understanding it conceptually, is better at times.
If you want the equation, here you go:
Vf=Vi+gt
Where
Vf=0m/s
Vi=7.5m/s
g=-9.8m/s^2
and
t=??
0=7.5m/s-9.8m/s^2*t
-7.5m/s/-9.8m/s^2=t
t=0.765
15m/s*Sin 30=7.5m/s
The deceleration due to gravity is 9.8m/s. So, after 1s the ball will be falling back down to the ground, which eliminates the first two answer choices. Does it take a half a second for the ball's velocity to reach 0m/s, or does it take slightly more time for the ball's velocity to reach 0m/s? In half a second, the ball will decelerate by 4.8m/s, which means that it will still have a velocity of 2.7m/s. The best answer choice is 0.77s.
****You could have just used an equation, but understanding it conceptually, is better at times.
If you want the equation, here you go:
Vf=Vi+gt
Where
Vf=0m/s
Vi=7.5m/s
g=-9.8m/s^2
and
t=??
0=7.5m/s-9.8m/s^2*t
-7.5m/s/-9.8m/s^2=t
t=0.765
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