a ball is thrown upward from the top of 25m building.initial speed is 12m/s.?

Question

at the same time a man is running on the ground at a distance of 31m from building. what must be the average speed of the person if he is to catch the ball

Answer 1

Assuming the ball is throw vertically upwards, then the vertical distance y is given by

y = 12t - 0.5gt^2 where g = 9.8 m/s^2

Find t when y= -25 leads to 4.9t^2 - 12t - 25 = 0. This leads to a solution (using quadratic formula) of t = 3.79 s

Therefore the man needs to cover 31 m in 3.79 s. His average speed needs to 31/3.79 or 8.17 m/s to catch the ball

Answer 2

I am assuming that this is a beginning physics question,...
Don't worry about the wind, or friction.
Also, the weight is irrelevant.

Once the ball leaves the hand of the person who threw it, it is accelerating at a speed of -9.8 m/s due to gravity. You have v-subscript0 as 12m/s, the acceleration, so you first need to calculate how far up the ball goes before it stops going up. Then you add that to the height of the building, and solve as though the ball was just dropped from that height (this time the initial velocity is 0).

Now calculate the amount of time it took for the ball to go up, and the amount of time it took to go all the way down and add the two numbers together.

That is how long the man has to run the 31 m. Just divide the 31m by the time that he has to get there.

Hope that helps.

Answer 3

The time it takes the ball to reach the ground must be equal to the time it takes the man to run from where he is to the point where he will be able to catch the ball.

First solve for the time, t, it takes the ball to reach its highest point. Then solve for the time, t', it takes the ball to reach the ground from its highest point. Total time, T, will then be t+t'.

To solve for the average speed, v, of the man use the basic formula,

D=vT

Where D is the distance given as 31m, v is the average speed in m/s, and T is the total time.

The applicable formula for solving t is:

s=ut+1/2at^2

where s is the distance in m from the top of the building to the highest point reached by the ball, u is the initial speed, a is the acceleration of gravity in m/s^2, and t is the time in seconds.

Given: u=0m/s, a=9.8m/s^2.

Why is u=0? Because we take the flight from the highest point to the top of the building. We know that at the highest point, the speed is equal to 0.

Substitute given values:

s=0*t+1/2*9.8*t^2

s=4.9t^2

We have 2 unknowns. So we solve for s first using the formula,

v^2-u^2=2as

In this case, for simplicity we take the speed at the highest point as the final speed, which we know is equal to 0.

Substitute given values:

0-12^2=2*(-9.8)s

a is negative because the flight of the ball is against gravity.

144=19.6s
s=144/19.6
=7.34m

Now substitute s in the above equation with 2 unknowns:

7.34=4.9t^2

t^2=7.34/4.9
=1.49
t=1.22 s

Solve for t' using the formula,

s=ut+1/2at^2

Where s is equal to 7.34m+25m=32.34m
u=0 (at the highest point)
a=9.8m/s^2

Substitute known values:

32.34=0*t+1/2*9.8*t^2
32.34=4.9t^2
t^2=32.34/4.9
=6.6
t'=2.57s

T=t+t'
=1.22+2.57
=3.79s

Now substitute the value of T and D in the formula,

D=vT
31=v*3.79
v=31/3.79
=8.17m/s

That speed is close to or even better than the world record for a 100m dash, if I'm not mistaken.

Note: In solving problems like this, take advantage of your knowledge that at the highest point of a projectile the vertical component of the speed is always equal to 0. (In this problem we are only dealing with a projectile without any horizontal speed component.)

Answer 4

You forgot that the ball that's coming down is already 15m. from the floor at time t = 0. For the ball going up: v1 = u1 - g * t (we subtract the acceleration * time from the preliminary velocity because of the fact the rigidity of gravity is working -against- the preliminary velocity) y1 = u1 * t - .5 * g * t ** 2 For the ball coming down: v2 = g * t y2 = 15 - .5 * g * t ** 2 (it starts at 15m. from the floor and gets closer by way of the years) you will then get a quadratic in t. between the values would be out of selection (the two unfavourable time, or the ball coming down has handed throughout the floor and is on its thank you to the Earth's middle).

Answer 5

To many variable. Weight of the ball. Wind speed. Trajectory of the ball. These would all play a part in order to make an accurate calculation

Answer 6

you should solve first this quadratic eqn:P
-25=-5t^2+12t
t=3.73s which is the time that man should travel 31m
V=31/3.73=8.31m/s average velocity