From the edge of a cliff, a gun fires a pellet straight upward. The pellet’s speed after leaving the gun is 30 m/s; the cliff is 75 meters tall. Assume that the ground is at height = 0 and that the positive direction is upward. At what time does the pellet hit the ground?
The correct answer and explanation is :
To determine the time at which the pellet hits the ground, we can analyze its motion in two phases: the ascent and the descent.
Ascent Phase:
- Initial velocity (v₀): 30 m/s (upward)
- Acceleration (a): -9.8 m/s² (due to gravity, acting downward)
- Final velocity at peak (v): 0 m/s (momentarily at the highest point)
Using the kinematic equation:
[ v = v₀ + at ]
Solving for time to reach the peak (t₁):
[ 0 = 30\, \text{m/s} – (9.8\, \text{m/s}²) \times t₁ ]
[ t₁ = \frac{30\, \text{m/s}}{9.8\, \text{m/s}²} \approx 3.06\, \text{seconds} ]
Descent Phase:
During descent, the pellet falls from the peak height back to the cliff’s edge and then continues to the ground.
- Height of the cliff (h): 75 meters
- Maximum height above the cliff (h₁):
Using the kinematic equation:
[ v² = v₀² + 2a s ]
Solving for s (maximum height above the cliff):
[ 0 = (30\, \text{m/s})² – 2 \times (9.8\, \text{m/s}²) \times s ]
[ s = \frac{(30\, \text{m/s})²}{2 \times 9.8\, \text{m/s}²} \approx 45.92\, \text{meters} ]
Total height from peak to ground (H):
[ H = h + h₁ = 75\, \text{meters} + 45.92\, \text{meters} = 120.92\, \text{meters} ]
Time to fall from peak to ground (t₂):
Using the kinematic equation:
[ s = v₀ t + \frac{1}{2} a t² ]
Here, initial velocity (v₀) is 0 (starting from rest at the peak), and acceleration (a) is 9.8 m/s² (downward).
[ 120.92\, \text{meters} = 0 + \frac{1}{2} \times 9.8\, \text{m/s}² \times t₂² ]
[ t₂² = \frac{120.92\, \text{meters} \times 2}{9.8\, \text{m/s}²} ]
[ t₂ \approx \sqrt{24.68} \approx 4.97\, \text{seconds} ]
Total Time to Hit the Ground (T):
[ T = t₁ + t₂ \approx 3.06\, \text{seconds} + 4.97\, \text{seconds} \approx 8.03\, \text{seconds} ]
Explanation:
When the pellet is fired upward from the edge of a 75-meter-high cliff with an initial velocity of 30 m/s, it first ascends to a maximum height of approximately 45.92 meters above the cliff. This ascent takes about 3.06 seconds. After reaching this peak, the pellet begins its descent, falling back past the cliff’s edge and continuing to the ground, covering a total distance of approximately 120.92 meters. The time taken for this descent is about 4.97 seconds. Therefore, the total time from firing to impact with the ground is approximately 8.03 seconds.
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