help with freefall physics problem please?

Question

A ball is thrown upward from the top of a 60.0 m tall building. The ball's initial speed is 12.0 m/s. At the same instant, a person is running on the ground at a distance of 41.0 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?

Answer 1

the person at the bottom can run around all day long and never encounter the ball at all.

according to your description of the problem, the ball is going up - you never mention an x vector (horizontal), only a y vector (vertical) to the ball's motion.

thus, the ball would land on the roof of the building... at the exact spot on the roof it was launched from.

Answer 2

Fun question. Is the person already running or does he/she have to accelerate from a standstill?

How exact do you want the answer? Off the top of my head I estimate that the ball will travel upwards for a little more than one second. Let's say it reaches a height of approx 71m.

Ignoring air resistance (of course)...

after the first second of freefall the ball will cover 9.8m
after the second second of freefall the ball covers 29.4m
after the third second, 58.8 meters
the last 13 meters are covered in about 1/4 second...lets give the time from 71m to ground approx 3.25 seconds

To cover 41m in the same time, your runner would have to be running over 12m/s constant velocity.

These are just estimates. I'd break out a calculator for more exact numbers.

Answer 3

First, find the time it takes for the ball to reach its apex:
v = a t
12 = 9.8t
t = 12/9.8
t = 1.2245sec

Now find the height the ball will acheive in this time:
s' = 1/2at^2 + s
s = 1/2 * 9.8m/ss * 1.2245s^2 + 60m
s = 4.9 * 1.4994 + 60
s = 67.347m

Now find the time it takes to fall to the ground from that height, using the same formula again.
67.347 = 1/2 * 9.8 t^2 + 0
134.6939 = 9.8t^2
t = sqrt(13.744)
t = 3.707sec.

Remember, it took another 1.2245sec for the ball going up, so it takes a total of 4.9318sec to hit the ground.

Now, just use the ol' distance = rate x time formula
41 = v * 4.9318
v = 41 / 4.9318
v = 8.313 meters per second. (for a diving catch right at the ground - lol)

Answer 4

First we need to work out the time taken for the ball to reach the ground for this we use the equation

S = UT +0.5(A)(T^2)

where S is displacement U is the initual speed and T is the time taken

it starts 60m up so S will be -60 taking up to be positve therefore acceleration under gravity = a = -9.8m/(s^2)

U = 12.0 m/s

T is our unknown

sub into eq

we get -60 = 12.0(T) + 0.5(-9.8)(T^2)

this will give you a quadratic in terms of T solve it using the quadradic formula, discounting your negitive answear for T

once you have the time it takes for the ball to reach the ground you know how long the runner has to cover that 41 m. so simply speed = distance/time

hey i wasnt going to do it all for you :P

EDIT: with referance to Christamud... in pyhics you disregaurd things like air resistance, building in the way...and logic

Answer 5

i would first determine how long it will take the ball to hit the ground after being thrown up in the air.

use the equation

d = .5at^2 + (v0)t + (d0)

d = final distance
a = acceleration = in this case -9.8 m/s^2
t = time
v0 = initial velocity = 12m/s
d0 = initial distance = 60m

you want to solve for final distance, d = 0, when it hits the ground

so

0 = -.5 * 9.8 * t^2 + 12t + 60
0 = -4.7t^2 + 12t + 60

use the quadratic equation to solve for t

t = 5.071s and -2.517s

the ball will hit the ground in 5.071sec

so the man must run 41m in 5.071sec

an avg velocity, v = distance/time = 41/5.071 = 7.88m/s
or

17.6mph

Answer 6

This is a questionably worded problem...if the ball were being launched directly upwards, wouldn't it just hit the top of the building on the way down? Or is the ball being launched at some angle?

Answer 7

listed below are the equations: a = consistent = -9.8 m/s² (in any respect cases) V = Vo + a*t Y = Yo + Vo*t + ½*a*t² (observed as 'place' in the question) Vo is +9.8 m/s and Yo is 0 on the precise of the development. placed the time values into the above equations and fill in the chart. The solutions to the two ?s below it is going to alter into obvious. the precise of the flight is while V = 0 and the roof point implies Y = 0