You shoot a rock with a wrist-rocket straight up, it takes 6 seconds for it to come back down (IGNORE THE HEIGHT YOU SHOT IT FROM.) How high did the rock go? what was the rock's speed initially going up (V1)?
How fast is it traveling when it hits the ground on the way down?
To determine the height the rock reached and its initial velocity (V1) when shot upwards, we can use the principles of projectile motion.
1. Finding the time it takes for the rock to reach its maximum height:
Since the rock is shot straight up and comes back down, the total time taken is 6 seconds. To find the time taken to reach the maximum height, we divide the total time by 2:
Time up = Time down = 6 seconds / 2 = 3 seconds
2. Finding the maximum height reached by the rock:
We can use the kinematic equation for vertical displacement (d = V1 * t + (1/2) * g * t^2) to find the maximum height. Since the rock starts at the height we shot it from, the initial displacement (d) is 0. Rearranging the equation, we get:
0 = V1 * 3 + (1/2) * (-9.8 m/s^2) * (3^2)
Simplifying, we have:
0 = 3V1 - 44.1
Rearranging further, we find:
V1 = 44.1 / 3 = 14.7 m/s
Therefore, the rock's initial speed (V1) when shot upwards is 14.7 m/s.
3. Finding the maximum height (H) reached by the rock:
Using the kinematic equation for vertical displacement again, we substitute the values:
H = 14.7 m/s * 3 s + (1/2) * (-9.8 m/s^2) * (3 s)^2
Simplifying, we have:
H = 44.1 m + (-44.1 m/2)
H = 44.1 m - 22.05 m
H = 22.05 m
Therefore, the rock reached a maximum height of 22.05 meters.
4. Finding the speed at which the rock hits the ground on the way down:
When the rock hits the ground, its velocity will be the same as its initial velocity but in the opposite direction. Hence, the speed at which the rock hits the ground is equal to the absolute value of the initial velocity (V1):
Speed = |V1| = |14.7 m/s| = 14.7 m/s
Therefore, the rock is traveling at a speed of 14.7 meters per second when it hits the ground on the way down.