The maximum speed of a child on a swing is 5.3 m/s. The child's height above the ground is 1.2 m at the lowest point in his motion. How high above the ground is he at his highest point? (Ignore dissipative forces).
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A child on a swing behaves like a body projected vertically upwards.
We therefore use the equations of motion.
Since we don't have time but we have initial and final velocity and we are looking for displacement s, we will use the equation below.
2as = v² - u²
Where a is the acceleration due to gravity, v is the final velocity and u the initial velocity.
We should take note that the projection is done from a point 1.2 m above the ground.
From our question:
u = 5.3m/s
v = 0
a = 10m/s²
2 × 10 × s = 0 - 5.3²
20s = - 28.09
S = 1.4045m
This is from point 1.2 m above the ground.
The highest point from the ground is thus :
1.2 m + 1.4045 m= 2.6045m
We therefore use the equations of motion.
Since we don't have time but we have initial and final velocity and we are looking for displacement s, we will use the equation below.
2as = v² - u²
Where a is the acceleration due to gravity, v is the final velocity and u the initial velocity.
We should take note that the projection is done from a point 1.2 m above the ground.
From our question:
u = 5.3m/s
v = 0
a = 10m/s²
2 × 10 × s = 0 - 5.3²
20s = - 28.09
S = 1.4045m
This is from point 1.2 m above the ground.
The highest point from the ground is thus :
1.2 m + 1.4045 m= 2.6045m
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