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
v = Vi - gt
when t = 2, v = 15
15 = Vi - 9.8*2
solve for Vi
then using that Vi solve for v at t = 4.2
a. is 34.6
so would b) be v=34.6-(9.8)(4.2)
To solve this problem, we can use the principles of kinematics by analyzing the relationships between position, velocity, and time. We'll break it down into two parts: the initial launch and the motion after 2.0 seconds.
(a) To find the speed at launch, we need to determine the initial velocity of the rock. We can use the equation:
v = u + at
Where:
v = final velocity (15 m/s)
u = initial velocity (unknown)
a = acceleration (which is due to gravity and can be calculated as -9.8 m/s²)
t = time (2.0 s)
Rearranging the equation to solve for the initial velocity (u), we get:
u = v - at
u = 15 m/s - (-9.8 m/s²)(2.0 s)
Now we plug in the given values:
u = 15 m/s + (19.6 m/s²)
u = 34.6 m/s
Therefore, the speed at launch is 34.6 m/s.
(b) To find the speed 4.2 seconds after launch, we will consider the free fall motion of the rock. During free fall, the only force acting on the rock is gravity, which causes it to accelerate at -9.8 m/s². We can use the equation:
v = u + at
Where:
v = final velocity (unknown)
u = initial velocity (34.6 m/s)
a = acceleration (-9.8 m/s²)
t = time (4.2 s)
Rearranging the equation to solve for the final velocity (v), we get:
v = u + at
v = 34.6 m/s + (-9.8 m/s²)(4.2 s)
Now we plug in the given values:
v = 34.6 m/s + (-41.16 m/s)
v = -6.56 m/s
Therefore, the speed 4.2 seconds after launch is -6.56 m/s.
Note: The negative sign indicates that the rock is moving downward at this point in time.