Question

While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 7.83 m/s. the stone subsequently falls to the ground, which is 10.3 m below the point where the stone leaves your hand. at what speed does the stone impact the ground? how much time is the stone in the air? ignore air resistance and take g = 9.81 m/s2. (this is not a suggestion to carry out such an experiment!)?

Answer 1

Initial velocity U = 7.83 Distance between hand and land s = 10.3 m; g = 9.81 m/s^2 We have V = U + at => V = U + gt => t = (V - U) / g
We have V^2 = U^2 + 2as = 7.83^2 + (2 x 9.81 x 10.3)
=> V^2 = 263.396 => V = 16.23
Now t = (16.23 - 7.83) / 9.81 => t1 = 0.856 s
At the drop V = 0 and gravity is against it g = -g
So V = U - gt=> 0 = U - gt => t = 7.83 / 9.81 => t2 = 0.798 s
t3 the time for the flight will be the same t3 = 0.798 s
Now the time taken by stone in the air = t1 + t2 + t3 = 2.45 s

Answer 2

a) For upward projection.
v = u -gt
but v = 0, thus u = gt,
t = u/g, t= 7.83/9.81, t = 0.798 sec
Therefore, distance, s
S = ut -1/2gt²
= 7.83(0.798) - 1/2(9.81)(0.798²)
= 6.248 - 3.124
= 3.124 m
Total Height is 10.3 + 3.124 = 13.424 m
For the case of free fall
s= ut + 1/2 gt² but u =0
s = 1/2 gt²
t² = (2 × 13.424)/9.81
t² = 2.7368
t = 1.654 s
Therefore, the time the stone remains in air will be 1.654 + 0.798
= 2.452 seconds

b) Speed of the impact on the ground
v= u + gt but, u = 0
thus v = gt
therefore, v = 9.81 × 1.653
= 16.216 m/s