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2 years ago

A ball is thrown up with a speed of 15m/s. How high will it go before it begins to fall? ( g = 10m/s2 )

Physics

1 answer:

LekaFEV [45]2 years ago

4 0

Answer:

Height is 11.25m

Explanation:

Given the following data;

Initial velocity, u = 0

Final velocity, v = 15m/s

Acceleration due to gravity, g = 10m/s²

To find the height, we would use the third equation of motion;

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

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement (height) measured in meters.

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

Making S the subject, we have;

$S = \frac {V^{2} - U^{2}}{2a}$

But a = g = 10m/s²

Substituting into the equation, we have;

$S = \frac {15^{2} - 0^{2}}{2*10}$

$S = \frac {225 - 0}{20}$

$S = \frac {225}{20}$

S = 11.25m

Therefore, the ball will reach a height of 11.25m before it begins to fall.

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