Answer :
Answer:
Maximum height, h = 1.27 meters
Explanation:
It is given that,
Initial speed of the basketball, u = 5 m/s
At maximum height, the final speed of the basketball, v = 0
Let h is the maximum height reached by the basketball from its release point. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2as[/tex]
Here, a = -g
[tex]-u^2=-2gh[/tex]
[tex]u^2=2gh[/tex]
[tex]h=\dfrac{u^2}{2g}[/tex]
[tex]h=\dfrac{(5)^2}{2\times 9.8}[/tex]
h = 1.27 meters
So, the maximum height reached by the basketball from its release point is 1.27 meters. Hence, this is the required solution.
The maximum height reached by the basket ball is 1.276 m.
The given parameters;
initial velocity of the basket ball, v = 5 m/s
The maximum height reached by the ball is calculated as;
[tex]v_f^2 = v_0^2 -2gh\\\\[/tex]
where;
- Vf is the final velocity at the maximum height = o
- Vo is the initial velocity
- h is the maximum height reached
The maximum height reached by the basket ball is calculate das;
[tex]v_f^2 = v_0^2 -2gh\\\\0 = v_0^2 -2gh\\\\2gh = v_0^2 \\\\h = \frac{v_0^2}{2g} \\\\ h = \frac{(5)^2}{2(9.8)} = 1.276 \ m[/tex]
Thus, the maximum height reached by the basket ball is 1.276 m.
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