Before working this problem, review Conceptual Example 15 . A pellet gun is fired straight downward from the edge of a cliff that is 22.3 m above the ground. The pellet strikes the ground with a speed of 34.9 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward?
h=v₀t +gt²/2
v= v₀+gt
v₀ =v-gt
h=( v-gt)t + gt²/2 =
=vt - gt² + gt²/2=
= vt - gt²/2.
gt² -2vt +2h = 0
t = (2•34.9±sqrt{(2•34.9)²-4•9.8•2•34.9}/2•9.8 = 0.7 s
v₀ =v-gt =34.9 – 9.8•0.7 = 28 m/s
Upward motion
H=v₀²/2g= 28²/2•9.8 = 40 m
To solve this problem, we can use the concept of projectile motion and the fact that the initial and final velocities are the same.
Step 1: Determine the time of flight for the pellet when fired downward.
We know that the initial velocity is 0 m/s (since the pellet starts from rest when fired downward) and the final velocity is -34.9 m/s (negative since it is moving downward). The acceleration due to gravity is -9.8 m/s^2 (negative because it acts downward).
Using the formula:
Final velocity = Initial velocity + (acceleration * time)
-34.9 m/s = 0 m/s - (9.8 m/s^2 * time)
Solving for time:
time = -34.9 m/s / -9.8 m/s^2
time = 3.56 s
Step 2: Calculate the height above the cliff edge when the pellet was fired downward.
Using the formula for vertical displacement:
Displacement = Initial velocity * time + (1/2) * acceleration * time^2
Since the initial velocity is 0 m/s, the equation simplifies to:
Displacement = (1/2) * acceleration * time^2
Plugging in the values:
Displacement = (1/2) * (-9.8 m/s^2) * (3.56 s)^2
Displacement = -62.1 m
Step 3: Calculate the distance above the cliff edge when the pellet is fired upward.
Since the pellet has to reach the same height above the cliff as it did when fired downward, the displacement when fired upward will be positive 62.1 m.
Using the same formula:
Displacement = Initial velocity * time + (1/2) * acceleration * time^2
We want to find the initial velocity when the pellet is fired upward. The acceleration due to gravity remains -9.8 m/s^2.
Plugging in the values:
62.1 m = Initial velocity * 3.56 s + (1/2) * (-9.8 m/s^2) * (3.56 s)^2
Solving for Initial velocity:
Initial velocity = (62.1 m - (1/2) * (-9.8 m/s^2) * (3.56 s)^2) / 3.56 s
Initial velocity = 62.1 m / 3.56 s
Initial velocity = 17.4 m/s
Therefore, the pellet would have gone approximately 17.4 m above the cliff edge if it was fired straight upward.
To solve this problem, we can use the concept of projectile motion. We know that the pellet is initially at a height of 22.3 m above the ground and it strikes the ground with a speed of 34.9 m/s. We need to find the maximum height the pellet would have reached if the gun had been fired straight upward.
Step 1: Understand the problem
In this problem, we have a projectile motion situation where the pellet is launched vertically upward. We are asked to find the maximum height reached by the pellet. We are given the initial velocity of the pellet (34.9 m/s) and the initial height of the pellet above the ground (22.3 m).
Step 2: Identify the knowns and unknowns
Known:
- Initial velocity (v₀) = 34.9 m/s (upward)
- Initial height (h₀) = 22.3 m
- Acceleration due to gravity (g) = -9.8 m/s² (downward)
Unknown:
- Maximum height reached (h_max)
Step 3: Choose equations and apply them
Since we are dealing with vertical motion, we can use the following kinematic equation:
v² = v₀² + 2aΔy
where:
- v = final velocity
- v₀ = initial velocity
- a = acceleration
- Δy = change in position (in this case, the change in height)
In this case, we need to find the maximum height (h_max), so Δy will be the change in height from the initial position to the highest point.
The final velocity (v) at the highest point is 0 since the pellet momentarily comes to rest before falling back down. Therefore, we can rewrite the equation as:
v₀² = 2aΔy
Plugging in the values we know:
(0 m/s)² = 2(-9.8 m/s²)Δy
Simplifying the equation:
0 = -19.6 Δy
Step 4: Solve for the unknown
Solving for Δy (change in height):
Δy = 0 m
This means the change in height (maximum height) is 0 meters. The pellet would not have gone any higher than the initial height of 22.3 m if the gun had been fired straight upward.
Step 5: Interpret the result
The result implies that the pellet, when fired straight upward, would reach its maximum height (0 m) at the same position where it was launched from (initial height of 22.3 m above the ground).