A stone with a mass of 2kg is thrown vertically upwards.it takes 5 seconds for the stone to reach the same height from which it was thrown. Determine th following:3.2.1 the volocity with which the stone was thrown

We can assume the rock was thrown from a height of 0.

So, now you know that
h(t) = vt-5t^2
h(5) = 0, so
5v-125 = 0
v = 25 m/s

note that the energy all together is the relationship between potential and kinetic energy

mgh=1/2mv2

v=u+at.. lest u is zero and a is 9.81 as t is 5. so v=49m/s
but note that by the law of conservation of energy above then make h the subject you get h=(1/2)(v2)/g.....we kno v and g ...so h=is 125 to three significant figures

To determine the velocity with which the stone was thrown, we can use the formula for velocity in vertical motion:

v = u + a * t

Where:
v = final velocity (in this case, the velocity when the stone reaches the same height from which it was thrown)
u = initial velocity (the velocity with which the stone was thrown)
a = acceleration (due to gravity, which is approximately 9.8 m/s²)
t = time taken to reach the final velocity

In this scenario, the stone is thrown vertically upwards, meaning the final velocity will be zero at the highest point. Therefore, we can rearrange the formula to solve for the initial velocity:

v = u + a * t
0 = u - 9.8 * 5

Simplifying the equation:

0 = u - 49

Rearranging to isolate the initial velocity, u:

u = 49

Therefore, the stone was thrown with an initial velocity of 49 m/s.