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 subs
equently 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!)?
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
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
At the bottom of the hill, the baby carriage will likely have less momentum Therefore, option D is correct. Solution: ... Therefore, at the bottom of the hill, the heavy truck will have more momentum and baby carriage will have less momentum.
