Question

50. A basketball player can jump 0.65 m off the ground. How
fast must the player be going when she leaves the ground?

Answer 1

The basketball player must be going at 3.57 m/s when she leaves the ground.

Given the following data:

• Height = 0.65 meter

We know that acceleration due to gravity (a) for an object or body is equal to 9.8 meter per seconds square.

To find how fast (velocity) the basketball player must be going when she leaves the ground, we would use the third equation of motion;

$V^2 = U^2 + 2aS$

Where:

• V is the final speed.

• U is the initial speed.

• a is the acceleration.

• S is the distance covered.

Substituting the given parameters into the formula, we have;

$V^2 = 0^2 + 2(9.8)(0.65)\\\\V^2 = 12.74\\\\V = \sqrt{12.74}$

Final velocity, V = 3.57 m/s

Therefore, the basketball player must be going at 3.57 m/s when she leaves the ground.

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