Question

11)

While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 5.71 m/s. The stone subsequently falls to the ground, which is 11.3 m below the point where the stone leaves his hand.

At what speed does the stone impact the ground? Ignore air resistance and use ?=9.81 m/s2 for the acceleration due to gravity.

impact speed:———-

How much time is the stone in the air?

elapsed time:———-

12)

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a large toy rocket to the back of a sled and take the modified sled to a large, flat, snowy field. You ignite the rocket and observe that the sled accelerates from rest in the forward direction at a rate of 12.5 m/s² for a time period of 3.50 s. After this time period, the rocket engine abruptly shuts off, and the sled subsequently undergoes a constant backward acceleration due to friction of 4.15 m/s².

After the rocket turns off, how much time does it take for the sled to come to a stop?

time:———

By the time the sled finally comes to a rest, how far has it traveled from its starting point?

distance traveled:———-

Answer 1

11)

Given:

$u=5.71\ \text{m/s}\\ h_1=11.3\ \text m\\ g=9.81\ \text{m/s}^2$

a)

Maximum height reached by the stone from the launching point is,

$h_2=\frac{u^2}{2g}\\ =\frac{5.71^2}{2\times9.81}\ \text m\\ =1.66\ \text m$

Then, final velocity of the stone is,

$v=\sqrt{2g(h_1+h_2)}\\ =\sqrt{2\times9.81(11.3+1.66)}\ \text {m/s}\\ =15.9\ \text{m/s}$

b)

Time to reach highest point is,

$t_1=\frac{u}{g}\\ =\frac{5.71}{9.81}\ \text s\\ =0.582\ \text s$

Time to return to ground from the highest point is,

$t_2=\sqrt{\frac{2(h_1+h_2)}{g}}\\ =\sqrt{\frac{2\times(11.3+1.66)}{9.81}}\ \text s\\ =1.63\ \text s$

Then, total time elapsed is,

$t=t_1+t_2\\ =0.582+1.63\ \text s\\ =2.21\ \text s$