A stone is thrown upward to just reach the top of a 50m office building.(a) calculate the initial speed that is required (b) how long does it take for the stone to return to the initial point of projection

Answer 1

To solve this problem, we can use the equation of motion for vertical displacement:

h = u*t + (1/2)*g*t^2

where h is the vertical displacement, u is the initial velocity, t is the time, and g is the acceleration due to gravity.

(a) In order for the stone to reach the top of the building, its vertical displacement must be equal to the height of the building, which is 50m. So we have:

50m = u*t - (1/2)*g*t^2

At the maximum height, the stone's vertical displacement is zero. So setting h = 0, we have:

0 = u*t - (1/2)*g*t^2

Solving this quadratic equation for t:

(1/2)*g*t^2 - u*t = 0

t * (0.5g*t - u) = 0

Either t = 0 (which is not the time we're interested in) or:

0.5g*t - u = 0
u = 0.5g*t

We know that the acceleration due to gravity, g, is approximately 9.8 m/s^2. Substituting this value and t = time of flight/2 into the equation, we can calculate the initial speed required.

(b) To calculate the time it takes for the stone to return to the initial point of projection, we need to find the time of flight, which is double the time it takes to reach the maximum height. From the previous equation, we have:

t = 2*u/g

Substituting the value of u from equation (a):

t = 2*(0.5g*t)/g
t = 1s

Therefore, it would take 1 second for the stone to return to the initial point of projection.