A basketball is thrown up into the air. It is released with an initial velocity of 8.5 m/s. How long does it take to get to the
top of its motion?
2 answers:
Answer:
0.87 s
Explanation:
initial velocity, u = 8.5 m/s
Let it takes time t to reach to maximum height. At maximum height the velocity is zero, so, v = 0
Use first equation of motion
v = u - gt
where, g be the acceleration due to gravity
0 = 8.5 - 9.8 t
t = 0.87 s
Thus, the time taken to reach at top is 0.87 s.
<h2>
It takes 0.867 seconds to get to the top of its motion</h2>
Explanation:
We have equation of motion v = u + at
Initial velocity, u = 8.5 m/s
Final velocity, v = 0 m/s - At maximum height
Time, t = ?
Acceleration , a = -9.81 m/s²
Substituting
v = u + at
0 = 8.5 + -9.81 x t
t = 0.867 s
It takes 0.867 seconds to get to the top of its motion
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