Question

A dynamite blast at a quarry launches a rock straight upward, and 1.7 s later it is rising at a rate of 17 m/s. Assuming air resistance has no effect on the rock, calculate its speed (a) at launch and (b) 4.9 s after launch.

Answer 1

Answer:

(a) $v_0= 34m/s$

(b) $v=14m/s$

Explanation:

We can derive the initial speed of the rock from the equation of the speed in function of the time:

$v=v_0-gt\\\\\implies v_0=v+gt$

Using the given values for the speed at time t=1.7s, we get:

$v_0=17m/s+(9.8m/s^{2})(1.7s)=34m/s$

In words, the speed of the rock at launch is 34m/s (a).

Next, we use this to calculate the speed at t=4.9s:

$v=v_0-gt\\\\v= 34m/s-(9.8m/s^{2})(4.9s)=-14m/s$

This means that the speed of the rock at 4.9s after the launch is 14m/s (b), and the negative sign means that it is moving downwards.