A dynamite blast at a quarry launches a rock straight upward, and 2.0 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)4.2 s after launch.
i found the the speed for (a) of when it was launch.. i calculate it from the formula .. v(t)= V+g(t)
V=15+9.8(2)=34.6
but how do i calculate it for (b) i tried plugging in 4.2 but i didn't work
how do i find the velocity 4.2s after lunch
To solve this problem, we can use the laws of motion, specifically the equations of motion for vertical motion. Let's go through the steps to find the answers to parts (a) and (b).
(a) To calculate the speed of the rock at launch, we need to use the equation for instantaneous velocity:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the rock is launched straight upward, so its final velocity at launch is 0 m/s (since it momentarily stops at its peak before falling). We are given the time as 2.0 s, and we need to find the initial velocity.
Using the equation, we have:
0 = u + (-9.8 m/s^2) x 2.0 s
Rearranging the equation, we get:
u = -(-9.8 m/s^2) x 2.0 s
u = 19.6 m/s
Therefore, the speed of the rock at launch is 19.6 m/s (upward).
(b) To calculate the speed of the rock 4.2 s after launch, we need to find the final velocity. Again, using the equation for instantaneous velocity:
v = u + at
We know the initial velocity u (which is 19.6 m/s), the acceleration a (-9.8 m/s^2 due to gravity), and the time t (4.2 s). We can now calculate the final velocity.
v = 19.6 m/s + (-9.8 m/s^2) x 4.2 s
v = 19.6 m/s - 41.16 m/s
v = -21.56 m/s
Therefore, the speed of the rock 4.2 s after launch is 21.56 m/s downward (negative velocity indicates downward direction).
Note: It is important to consider the negative sign while interpreting the direction (upward or downward) of the rock's motion.