Question

A dynamite blast at a quarry launches a rock straight upward, and 2.0 s later it is rising at a rate of 13 m/s. Assuming air resistance has no effect on the rock, calculate its speed (a) at launch and (b) 4.6 s after launch.(a) Number ________Units ________(b) Number ________Units ________

Answer 1

Given that the rock is moving upward, so it will have an acceleration due to gravity, g = -9.81 m/s^2

When time, t =2 seconds, the velocity is v = 13m/s.

We have to find (a) speed at the time of launch

(b) speed of rock when time , t' =4.6 s

(a) Let the launch speed be u.

Also, the displacement of the rock after 2 s will be

`\begin{gathered} s=v* t \\ =13*2\text{ m} \\ =26\text{ m} \end{gathered}`

Using the equation of motion,

`s=ut-(1)/(2)gt^2`

The launch speed can be calculated as

`u=(s+(1)/(2)gt^2)/(t)`

Substituting the values, speed will be

`\begin{gathered} u=(26+(1)/(2)*9.81*(2)^2)/(2) \\ =22.81\text{ m/s} \end{gathered}`

Thus, the launch speed is 22.81 m/s

(b) The time is t'=4.6 s.

Let the speed of the rock at this time be v'.

The speed can be calculated by the formula,

`v^(\prime)=u-gt`

Substituting the values, the speed will be

`\begin{gathered} v^(\prime)=22.81-9.81*4.6 \\ =-22.316\text{ m/s} \end{gathered}`

Here, the negative sign of speed indicates that the rock is moving downwards at this time.

The magnitude of speed at t'=4.6 s is 22.316 m/s