With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 80 m?

Question

How long will it be in the air?

Answer 1

so we know that y=80 and since it is maximum height v=0
let us choose that the ball the thrown at t=0 and at a height of y=0 so from the equations v^2=v0^2+2a x we have
0=v0^2+2(80)(-9.8)
so solve for v0 we get v0=39.6 m/s
then we will us v=v0+at
this gives us 0=39.6-9.8*t
so we get time to get to max height is 4.04 seconds
But by symmetry we now the time to get up to max height is the same as to get back down from max height
so the time in the air is 8.08 seconds

Answer 2

Turn the problem around: how fast would a ball be travelling if dropped from a height of 80m?

The ball will be decelerated going upwards at 9.8m/s^2 just like it would be accelerated goind down at the same rate.

80 = 9.8 * t^2...now solve for t in seconds.

Answer 3

People make it so hard. Its easy. Kinetic energy at launch = potential energy at max height.

So 1/2 mv^2 = mgh

m cancels - so you do not need to know it at all

v = sqrt (2gh) = sqrt (2.10.80) = sqrt 1600 = 40 m/s

Answer 4

The total time for the return trip will be 8.08 seconds. The initial velocity will have to be 79.2 metres per second.

Answer 5

lol again you need mass of the ball

but interesting questions