Question

A dynamite blast at a quarry launches a chunk of rock straight upward, and 2.0s later it is risinv at a speed of 15m/s. Assuming air resistance has no effect on the rock, calculate its speed (a) at launch and (b) 5.0s after launch

Answer 1

(a) To solve for the initial speed of the rock at launch, we can use the kinematic equation:

v = v0 + at

Where:

v = final velocity (15m/s)

v0 = initial velocity (what we're solving for)

a = acceleration due to gravity (-9.8m/s^2)

t = time (2.0s)

Plugging in the values, we get:

15m/s = v0 - 9.8m/s^2 (2.0s)

v0 = 34.6m/s

Therefore, the initial speed of the rock at launch was approximately 34.6m/s.

(b) To solve for the speed of the rock 5.0s after launch, we can use the same kinematic equation:

v = v0 + at

But this time, we need to add the additional time and distance that the rock traveled after the initial 2.0s. To do this, we'll use the equation:

d = v0t + 1/2at^2

Where:

d = distance traveled

v0 = initial velocity (34.6m/s)

a = acceleration due to gravity (-9.8m/s^2)

t = time (5.0s - 2.0s = 3.0s)

Plugging in the values, we get:

d = (34.6m/s)(3.0s) + 1/2(-9.8m/s^2)(3.0s)^2

d = 103.8m - 44.1m

d = 59.7m

So, the rock traveled 59.7m in the additional 3.0s after the initial 2.0s. Now we can find its speed using the kinematic equation:

v = v0 + at

Where:

v0 = final velocity from before (15m/s)

a = acceleration due to gravity (-9.8m/s^2)

t = time (3.0s)

Plugging in the values, we get:

v = 15m/s - 9.8m/s^2 (3.0s)

v = -12.6m/s

Note that the velocity is negative because the rock is now moving downward. Therefore, the speed of the rock 5.0s after launch is approximately 12.6m/s.