A dynamite blast at a quarry launches a rock straight upward, and 2.2 s later it is rising at a rate of 17 m/s. Assuming air resistance has no effect on the rock, calculate its speed (a) at launch and (b)4.4 s after launch.
a. V = Vo + g*t = 17 m/s.
Vo = 17-g*t = 17-(-9.8*2.2) = 38.56 m/s.
b. V = 38.56 -9.8*4.4 = -4.56 m/s.
The negative sign means the rock is on its' way down.
To solve this problem, we can use the following kinematic equations:
1. v = u + gt
2. s = ut + 1/2gt^2
where:
v = final velocity
u = initial velocity
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
s = distance
(a) To calculate the rock's speed at launch, we can use equation 1 and substitute the given values:
v = u + gt
17 m/s = u + 9.8 m/s^2 * 2.2 s
Rearranging the equation, we get:
u = 17 m/s - 9.8 m/s^2 * 2.2 s
u = 17 m/s - 21.56 m/s
u ≈ -4.56 m/s
So, the rock's speed at launch is approximately -4.56 m/s (since it's moving upward).
(b) To calculate the rock's speed 4.4 seconds after launch, we can use equation 1 again:
v = u + gt
v = -4.56 m/s + 9.8 m/s^2 * 4.4 s
v = -4.56 m/s + 43.12 m/s
v ≈ 38.56 m/s
So, the rock's speed 4.4 seconds after launch is approximately 38.56 m/s.
To solve this problem, we can use the equations of motion to calculate the speed of the rock at different times. The equations of motion are:
1. v = u + at
2. s = ut + (1/2)at^2
3. v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
s = displacement
Let's start by calculating the speed at launch (t = 0 s):
(a) Speed at launch (t = 0s):
Given:
u = ?
v = 17 m/s (at t = 2.2 s)
a = -9.8 m/s^2 (assuming positive direction is upward)
Using Equation 1, we can rearrange it to solve for the initial velocity (u):
v = u + at
17 m/s = u - (9.8 m/s^2)(2.2 s)
17 m/s = u - 21.56 m/s
u = 17 m/s + 21.56 m/s
u = 38.56 m/s
Therefore, the speed at launch (t = 0 s) is 38.56 m/s.
(b) Speed 4.4 seconds after launch (t = 4.4 s):
Given:
u = 38.56 m/s (at t = 0 s)
v = ?
a = -9.8 m/s^2 (assuming positive direction is upward)
t = 4.4 s
Using Equation 1 to calculate the speed:
v = u + at
v = 38.56 m/s - (9.8 m/s^2)(4.4 s)
v = 38.56 m/s - 43.12 m/s
v = -4.56 m/s
Therefore, 4.4 seconds after launch (t = 4.4 s), the speed of the rock is -4.56 m/s. The negative sign indicates that the rock is moving downward.