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energy conservation-problems 1. A slingshot fires a pebble from the top of a building at a speed of 14.0m/is. The building is 31.0m tall. Ig
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energy conservation-problems 1. A slingshot fires a pebble from the top of a building at a speed of 14.0m/is. The building is 31.0m tall. Ignoring all frictional effects, find the speed with which the pebble strikes the ground
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Physics
3 years
2021-09-02T12:16:13+00:00
2021-09-02T12:16:13+00:00 1 Answers
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Answers ( )
Answer:
Horizontal velocity is 14 m/s
Vertical velocity is 28.3 m/s
Explanation:
Hello dear friend, you have not mentioned the type of speed that you require for this problem.
In this case, there are three possibilities with which the slingshot can be fired i.e. Horizontally, Vertically straight up and vertically straight down. Below is the explanation / answer to all three possibilities
Fired horizontally:
initial conditions:
Vertical Velocity = 0 ; Horizontal Velocity = 14m/s
final conditions:
Vertical Velocity (v² = u² + 2gs) but initial vertical velocity is zero
v² = 2gs so v² = 2(9.8)(31) = 607
v = 24.6m/s
but Horizontal Velocity is still = 14m/s
Resultant velocity from these two velocity components (Pythagoras theorem)
V² = (v horizontal)² + (v vertical)² = 14² + 24.6²
V = 28.3m/s
angle = tan ⁻¹(24.3/14) = 60.1⁰
V = 28.3m/s at angle of 60.1⁰ to the horizontal
Fired Vertically Straight Up
distance before the pebble reaches maximum height from top of building
v² = u² + 2gs
where, v is zero at maximum height
g is minus for upward motion.
v² = u² + 2gs
0 = 14² – 2(9.8)s
s = 196/19.6 = 10.0m
totals distance from maximum height to the ground = 10.0 m + 31.0 m = 41.0m
v² = u² + 2gs
now u from maximum height is 0 and g is positive for downward motion
v² = 2gs
v² = 2(9.8)(41.0)
v = 28.3m/s
v = 28.3m/s vertically straight up
Fired Vertically Straight Down
v² = u² + 2gs
u = 14m/s, g = 9.8m/s², s = 31.0m
v² = 14² + 2(9.8)(31.0)
v = 28.3m/s
v = 28.3m/s vertically straight down