Question

a dynamite blast at a quarry launches a chunk of rock straight upward, and 3.8 s later it is rising at a speed of 13.1 m/s. assuming air resistance has no effect on the rock, calculate its speed at launch.

Answer 1

The initial speed of the rock at launch was approximately 50.34 m/s, calculated using the kinematic equation v = u + at, taking gravity as -9.8 m/s^2.

Step-by-step explanation:

We can calculate the speed of the rock at launch by using the kinematic equations for uniformly accelerated motion, specifically the one that relates the final velocity (v), initial velocity (u), acceleration (a), and time (t):

v = u + at

For an object thrown upwards, the acceleration due to gravity (g) is -9.8 m/s2 (negative because it is directed downwards). Since 3.8 seconds later the rock is still rising, it means it has not yet reached the peak of its ascent. We are given the final velocity (v = 13.1 m/s) and time (t = 3.8 s).

Substitute a with -g and rearrange the equation to solve for u:

u = v - at = 13.1 m/s - (-9.8 m/s2)(3.8 s)

Perform the calculation for u:

u = 13.1 m/s + (9.8 m/s2)(3.8 s)

u = 13.1 m/s + 37.24 m/s

u = 50.34 m/s

Therefore, the initial speed of the rock at launch was approximately 50.34 m/s.