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Sagot :
Given:
• Time, t =5 seconds
,• Initial velocity, u = 15 m/s
,• Final velocity, v = 34 m/s
Given that the ball is projected upwards from a bridge, let's find the height of the bridge.
Let's first find the height from the top of the bridge to the maximum height.
To find the height apply the formula:
[tex]v^2=u^2-2as[/tex]Where:
v is the final velocity = 0 m/s (velocity at max height).
u is the initial velocity = 15 m/s
a is acceleration due to gravity = 9.8 m/s²
s is the height.
[tex]\begin{gathered} 0^2=15^2-2(9.8)s \\ \\ 0=225-19.6s \\ \\ 19.6s=225 \\ \\ s=\frac{225}{19.6} \\ \\ s=11.48\text{ m} \end{gathered}[/tex]The distance from the bridge to the maximum height is 11.48 m.
Now, let's find the maximum height to the ground.
[tex]v^2=u^2+2as[/tex]Where:
v is the final velocity = 34 m/s
u is the initial velocity = 0 m/s (velocity at the top).
a is acceleration due to gravity = 9.8 m/s²
s is the maximum height.
Thus, we have:
[tex]\begin{gathered} 34^2=0^2+2(9.8)s \\ \\ 1156=19.6s \\ \\ s=\frac{1156}{19.6} \\ \\ s=58.9\approx59\text{ m} \end{gathered}[/tex]The maximum height is 59 meters.
To find the height of the bridge, we have:
Height of bridge = Maximum height - Distance from top of bridge to max height.
Height of bridge = 59m - 11.48m = 47.52 m
Therefore, the height of the bridge is 47.52 meters.
• ANSWER:
47.52 meters.
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