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A basketball is thrown straight up in the air at 15 m/s how high does it go?

User Cbartel
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Final answer:

To determine the maximum height reached by a basketball thrown straight up at 15 m/s, one should use the equation of motion under constant acceleration due to gravity. By applying the formula v^2 = u^2 + 2as, with v as final velocity (0 m/s at the peak), u as initial velocity, a as the acceleration (-9.81 m/s^2), the maximum height s can be calculated.

Step-by-step explanation:

To find out how high the basketball goes when it's thrown straight up in the air at 15 m/s, we need to apply the equations of motion under constant acceleration. In this case, the acceleration is due to gravity, which we'll take as 9.81 m/s2 (negative because it is acting downwards).

Using the formula for vertical motion v2 = u2 + 2as, where v is the final velocity (0 m/s at the highest point), u is the initial velocity (15 m/s), a is the acceleration due to gravity (-9.81 m/s2), and s is the distance (height), we can solve for s.

Putting in the values we get 0 = (15)2 + 2*(-9.81)*s, which simplifies to s = (15)2 / (2*9.81). Calculating this gives us the maximum height the basketball reaches. Remember, in projectile motion, the initial velocity plays a crucial role in determining the final height and hang time of the projectile.

User Alexander Tsepkov
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