with what velocity must a ball be thrown vertically upwards to reach a height of 25m? find the time it took?

Question

provide me with the necessary formula and why you chose that one. gracias.

Answer 1

The speed you want is the same as the reached by a ball dropped from 25 m.

The simplest way to do this is:
1. Figure out how long the ball will fall.
That time is sqrt(25/9.8).
This is based on the formula:
distance = acceleration x time^2
In this case,
25 = 9.8 t^2 (where 9.8 is the acceleration, caused by gravity, of 9.8 meters per sec^2).
When you solve for time, you get t = sqrt(25/9.8) = 1.6 sec

2. Figure out how fast a falling object is going at the end of 1.6 sec.
Answer: 1.6 sec x 9.8 m/sec^2 = 15.7 m/sec

Those calculations are for an object falling from 25 m, but if you reverse the process, you just throw the ball upward at 15.7 m/sec, it then decelerates at 9.8 m/sec^2, and reaches a velocity of zero after 1.6 sec. at a height of 25 m.

Answer 2

The easiest way to solve this problem is to use the equation of motion v = sqrt (2 x a x h) = sqrt ( 2 x 9.8 x 25) = 22.136 m/s. The elapsed time can be found in either of two ways, but the easiest is to use vf = vi + a x t where vf = 0 at 25 m peak height, vi = 22.136 m/s, a = 9.8 m/s^2. Solve for t = (vf - vi)/a or t = 2.259 s.

Answer 3

Vf^2 = Vo^2 + 2gs

I get 22.136 m/sec

For the time (to the peak height of 25 m)

Vf = Vo - gt

I get t = 2.259 sec.

[ Now you can ck these values in the distance formula:
s = Vot + (1/2)g t^2 keeping in mind that if you choose Vo positive then g needs to be negative.]

Answer 4

Vf^2 = 2ad + Vo^2
because we are looking for inital velocity and given by the height

after finding the initial velocity
use, Vf = at+Vo