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?

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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Scottsdale, AZ

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.

Blacksburg, VA

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"

Apison, TN

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 + V_{0}. In this case the initial velocity V_{0} is 50m/sec. At its apex the velocity has become zero (0). The acceleration is gravity a = -9.8m/sec^{2}. Which gives us,

0 = (-9.8m/sec^{2})t + 50m/sec

(9.8m/sec^{2})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)at^{2} + V_{0}t + d_{0}

Initial distance (d_{0}) is zero (0).

Initial velocity is 50.0 m/s.

Acceleration is gravity -9.8 m/s^{2}.

And the time it took to reach its maximum height is 5.10 sec.

So we get:

d = (1/2)(-9.8 m/s^{2})(5.10s)^{2} + (50.0m/s)(5.10s) + 0

d = (-4.9 m/s^{2})(26.03s^{2}) + (255.10m)

d = (-127.45m) + (255.10m)

d = 127.65 m

Chicago, IL

a) v_{0} - at = 0

50m/s - (9.8m/s^{2})t = 0

I will be omitting the units:

50 = 9.8t ---> t ≈ 5.1 seconds

b) v_{0 }t - (1/2)at^{2 }(height achieved)

50(5.1) - .5 * 9.8 * 5.1^{2} ≈ 127.55 meters

c) should be 50m/s also.

New Wilmington, PA

In this and the other free-fall problems you posted, you use the kinematic equations,

Δy=½at^{2}+v_{i}t and v_{f}=v_{i}+at with a=-g=-9.8 m/s²,

together with the fact that v=0 at the highest point.

So, let v_{i}=50 m/s, v_{f}=0 at the highest point.

a) From v_{f}=v_{i}+at you have

0=50-9.8t, t=50/9.8=5.10 s

b) Use Δy=½at^{2}+v_{i}t 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)