We have that the maximum height reached by the basketball from its release point is

From the question we are told
- A basketball is tossed upwards with a speed of 5.0 m/s. We can ignore air resistance.
- What is the maximum height reached by the basketball from its release point?
Generally the Newtons equation for Motion is mathematically given as


Therefore
The maximum height reached by the basketball from its release point is

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Jim walks 7 miles from the starting point to the end point
i guess so
Answer:F(of gravity) = MA
F(normal force) = MA * cos(angle)
F = 72 * 9.81 * cos28
Don't have a calculator, so can't really do all the math right there. So just plug that in
Explanation:
i dont really know
In order to compute the torque required, we may apply Newton's second law for circular motion:
Torque = moment of inertia * angular acceleration
For this, we require the angular acceleration, α. We may calculate this using:
α = Δω/Δt
The time taken to achieve rotational speed may be calculated using:
time = 1 revolution * 2π radians per revolution / 3.5 radians per second
time = 1.80 seconds
α = (3.5 - 0) / 1.8
α = 1.94 rad/s²
The moment of inertia of a thin disc is given by:
I = MR²/2
I = (0.21*0.1525²)/2
I = 0.002
τ = 1.94 * 0.002
τ = 0.004
The torque is 0.004