A pitcher stands 18m from home plate. If he/she throws the ball at 45 m/s, how long after the release will the ball cross home plate?
I was out sick, and my teacher refuses to actually TEACH, so I'm utterly confused and I have a makeup test TOMORROW. Please help!
Btw, I understand dimensional analysis completely. I don't know if I have to use it though.
Thanks!
I was out sick, and my teacher refuses to actually TEACH, so I'm utterly confused and I have a makeup test TOMORROW. Please help!
Btw, I understand dimensional analysis completely. I don't know if I have to use it though.
Thanks!
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How long to go 18 m @ 45 m/s?
d = v*t
18 m = 45 m/s * t
t = 18 m/45m/s (meters cancel and the seconds "flip" to the numerator because they are in the denominator UNDER a divide.)
t = 0.4 s
d = v*t
18 m = 45 m/s * t
t = 18 m/45m/s (meters cancel and the seconds "flip" to the numerator because they are in the denominator UNDER a divide.)
t = 0.4 s
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There are many ways to do this. Here's one. However, if you are not familiar with these sorts of calculations, watch the video in the link before going through my solution.
I'll use positive for upwards vectors, and negative for downwards vectors. Take g = -9.8m/s².
Suppose the ball is thrown at angle θ to the horizontal and is in the air for time T.
The horizontal component of velocity = 45cos(θ)
The initial vertical component of velocity = 45sin(θ)
The final vertical component of velocity = -45sin(θ)
(e.g if is was initially travelling upwards at 7m/s, it will finish travelling downwards at 7m/s (= -7m/s))
Considering vertical motion only, and using the standard equation v=u+at:
v=u +at
-45sin(θ) = 45sin(θ) + (-9.8)T
9.8T = 90sin(θ)
sin(θ) = 0.1089T
Horizontally the velocity is constant ,so (using distance = speed x time):
18 = 45cos(θ) x T
cos(θ) = 0.4/T
Since sin²()+cos²() = 1
(0.1089T)² + (0.4/T)² = 1
0.01186T² + 0.16/T² = 1
Let A = T²
0.01186A + 0.16/A = 1
0.01186A² + 0.16 = A (multiplying though by A)
0.01186A² - A + 0.16 =0
Solve the quadratic equation in the normal way (messy)
A = 84.16 or 0.1603
Since A = T², A = √
T = √84.16 = 9.2s or
T = √0.1603 = 0.40s
Both solutions are possible. 9.2s will correspond to the ball being thrown almost vertically upwards. 0.40s will correspond to the ball being thrown almost horizontally.
Assuming I have made no arithmetic mistakes!
I'll use positive for upwards vectors, and negative for downwards vectors. Take g = -9.8m/s².
Suppose the ball is thrown at angle θ to the horizontal and is in the air for time T.
The horizontal component of velocity = 45cos(θ)
The initial vertical component of velocity = 45sin(θ)
The final vertical component of velocity = -45sin(θ)
(e.g if is was initially travelling upwards at 7m/s, it will finish travelling downwards at 7m/s (= -7m/s))
Considering vertical motion only, and using the standard equation v=u+at:
v=u +at
-45sin(θ) = 45sin(θ) + (-9.8)T
9.8T = 90sin(θ)
sin(θ) = 0.1089T
Horizontally the velocity is constant ,so (using distance = speed x time):
18 = 45cos(θ) x T
cos(θ) = 0.4/T
Since sin²()+cos²() = 1
(0.1089T)² + (0.4/T)² = 1
0.01186T² + 0.16/T² = 1
Let A = T²
0.01186A + 0.16/A = 1
0.01186A² + 0.16 = A (multiplying though by A)
0.01186A² - A + 0.16 =0
Solve the quadratic equation in the normal way (messy)
A = 84.16 or 0.1603
Since A = T², A = √
T = √84.16 = 9.2s or
T = √0.1603 = 0.40s
Both solutions are possible. 9.2s will correspond to the ball being thrown almost vertically upwards. 0.40s will correspond to the ball being thrown almost horizontally.
Assuming I have made no arithmetic mistakes!
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If the pitcher through the ball at 18 m/s and was 18 m from the plate, it would take 1 second to get there right? 18m/18m/s=1s
But the pitcher is throwing it faster than that, so it will take less time to get there. 18m/45m/s = .4 s
But the pitcher is throwing it faster than that, so it will take less time to get there. 18m/45m/s = .4 s