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?

Tutors, sign in to answer this question.

Marked as Best Answer

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.

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

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.

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)

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.

Cindy B.

Math, science, and English tutor; K-12 and remedial

$11.25 per 15 min

View Profile >

Nick S.

Knowledgeable and willing to help with anything STEM related.

$10 per 15 min

View Profile >

Skyler H.

Specialized Techniques in Physics and Chemistry

$8.75 per 15 min

View Profile >