Question

A dynamite blast at a quarry launches a rock straight upward, and 2.4 s later it is rising at a rate of 15 m/s. Assuming air resistance has no effect on the rock, calculate its speed (a) at launch and (b) 4.6 s after launch.

Answer 1

After 2.4 s the rock is traveling at 15 m/s.

and

we know that rock having no acceleration of its own is only acted on by g,

therefore it had a (2.4 x g) = 2.4(9.81) = 23.5 m/s loss of speed at t = 2.4 s,

so

originally the initial Upward velocity of rock = 15 + 23.5 = 38.5 m/s ANS-a

which means it would have reached max height at: 38.5/g = 38.5/9.81 = 3.929 s

therefore

at 4.6 s after launch rock has fallen from its max height = 4.6 – 3.929 = 0.67 s

an object that falls for 0.67 s has a speed of:

V = gt = 9.81(0.67) = 6.582 m/s ANS-b