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How high does it go?

A ball is thrown up at velocity of 50.0 m/s. Consider no air resistance.
a) How much time does it take to reach its maximum height?
b) How high does it go?
c) How fast is it going when it reaches the ground again?

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Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
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a) v = gt
t = v/g = 50.0/9.8 = 5.1 sec
 
b) v^2 = 2gh
h = v^2/(2g) = 50.0^2/(2*9.8) = 128 m
 
c) 50.0 m/s, the same as the initial speed.
 
Brad M. | Professional Tutor: Ultra Streamlined Math - Physics - AccountingProfessional Tutor: Ultra Streamlined M...
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Sun -- here's the "play-by-play" ... gravity decels at 10m/s/s ==> 5s of "delta 10's" slows it to zero ... ave speed of 25m/s runs 5s => 125m ... Vf = Vo "what goes up must come down"
David F. | USNRC Training Professional – Mathematics TutorUSNRC Training Professional – Mathematic...
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The first thing to answer part c.  Since there is no air resistance then the same acceleration that slows the ball on the way up will also accelerate it on the way down.  So, it hits the ground at the same speed of 50.0 m/sec.
 
Part a is the next simplest.  V=at + V0.  In this case the initial velocity V0 is 50m/sec.  At its apex the velocity has become zero (0).  The acceleration is gravity a = -9.8m/sec2.  Which gives us,
 
0 = (-9.8m/sec2)t + 50m/sec
(9.8m/sec2)t = 50m/sec
t = (50/9.8) sec
t = 5.10 sec
 
Lastly for part b the formula is the standard distance formula: d = (1/2)at2 + V0t + d0
 
Initial distance (d0) is zero (0).
Initial velocity is 50.0 m/s.
Acceleration is gravity -9.8 m/s2.
And the time it took to reach its maximum height is 5.10 sec.
 
So we get:
 
d = (1/2)(-9.8 m/s2)(5.10s)2 + (50.0m/s)(5.10s) + 0
d = (-4.9 m/s2)(26.03s2) + (255.10m)
d = (-127.45m) + (255.10m)
d = 127.65 m
 
Ralph L. | Algebra I, II, Visual Basic, Beginning C++ tutorAlgebra I, II, Visual Basic, Beginning C...
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a)   v0 - at = 0  
      50m/s - (9.8m/s2)t = 0
 
       I will be omitting the units:
 
        50 = 9.8t   ---> t ≈ 5.1 seconds
 
b) v0 t - (1/2)at2 (height achieved)
 
50(5.1) - .5 * 9.8 * 5.12 ≈ 127.55 meters
 
c) should be 50m/s also.
 
 
Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
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In this and the other free-fall problems you posted, you use the kinematic equations,
 
Δy=½at2+vit  and vf=vi+at with a=-g=-9.8 m/s²,
 
together with the fact that v=0 at the highest point.
 
So, let vi=50 m/s, vf=0 at the highest point.
a) From vf=vi+at you have
0=50-9.8t,   t=50/9.8=5.10 s
 
b) Use Δy=½at2+vit with the t we found in a):
Δy=½(-9.8)(5.10)²+50(5.10)=128 m
 
c) Since it takes as long to go up as it takes to fall back down to its original position (5.10 s), the final velocity will be -50 m/s (down)