A dynamite blast at a quarry launches a chunk of rock straight upward, and 2.0s later it is rising at a speed of 15m/s. Assuming air resistance has no effect, calculate its speed A) at launch and B) 5.0s after lauch
v = Vi - 9.8 t
15 = Vi - 19.6
Vi = 34.6 s
so
v = 34.6 - 9.8 t
You can do that for t = 5
To calculate the speed of the rock at launch and 5.0 seconds after launch, we will use the principles of projectile motion.
Given:
Time since launch (t) = 2.0 s
Final velocity at 2.0 seconds (vf) = 15 m/s
A) To calculate the speed at launch:
The initial velocity (vi) can be found using the equation:
vf = vi + a * t
Since the rock is launched straight up, the acceleration (a) will be equal to -9.8 m/s^2 (acceleration due to gravity). Substituting the given values into the equation:
15 m/s = vi - 9.8 m/s^2 * 2.0 s
Rearranging the equation, we get:
vi = 15 m/s + 9.8 m/s^2 * 2.0 s
vi = 15 m/s + 19.6 m/s
vi = 34.6 m/s
Therefore, the speed of the rock at launch is 34.6 m/s.
B) To calculate the speed 5.0 seconds after launch:
We can use the equation of motion for vertical motion:
vf = vi + a * t
Here, we will assume that the rock is still moving upward 5.0 seconds after launch, so the final velocity (vf) will be positive. Rearranging the equation, we get:
vf = vi + a * t
vf = 34.6 m/s - 9.8 m/s^2 * 5.0 s
Calculating the expression above:
vf = 34.6 m/s - 49.0 m/s
vf = -14.4 m/s
Therefore, the speed of the rock 5.0 seconds after launch is 14.4 m/s downward.
To calculate the initial speed at launch of the rock, we'll use the fact that the acceleration due to gravity is constant. We'll also assume that the positive direction is upwards.
1. Calculate the acceleration:
Since the rock is moving upwards, and we'll assume that the positive direction is upwards, the acceleration due to gravity (g) will be negative. On Earth, the standard value for g is approximately -9.8 m/s^2.
2. Calculate the final velocity:
The velocity at a specific time can be calculated using the equation:
v = u + at
Where:
v is the final velocity (15 m/s),
u is the initial velocity (unknown in this case),
a is the acceleration (-9.8 m/s^2),
t is the time (2.0 seconds).
Rearranging the equation, we have:
u = v - at
Substituting the given values:
u = 15 m/s - (-9.8 m/s^2) * 2.0 s
Simplifying the equation, we get:
u = 15 m/s + 19.6 m/s
u = 34.6 m/s
So, the initial speed (or velocity) at launch is 34.6 m/s.
3. Calculate the speed 5.0 seconds after launch:
To find the speed 5.0 seconds after launch, we need to use the same equation:
v = u + at
Where:
v is the final velocity (unknown in this case),
u is the initial velocity (34.6 m/s),
a is the acceleration (-9.8 m/s^2),
t is the time (5.0 seconds).
Substituting the given values, we have:
v = 34.6 m/s - (9.8 m/s^2) * 5.0 s
Simplifying the equation, we get:
v = 34.6 m/s - 49 m/s
v = -14.4 m/s
The negative sign indicates that the rock is moving downwards. So, the speed 5.0 seconds after launch is 14.4 m/s downwards.