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2 years ago

11

A ball is thrown up into the air with an initial velocity of 18 m/s. A) How high does the ball go? B) Calculate the time needed

for the ball to reach its max height.

1 answer:

kaheart [24]2 years ago

7 0

Answer:

B) t = 1.83 [s]

A) y = 16.51 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f} =v_{o} -g*t$

where:

Vf = final velocity = 0

Vo = initial velocity = 18 [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time [s]

Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.

A) The maximum height is reached when the final velocity of the ball is zero.

0 = 18 - (9.81*t)

9.81*t = 18

t = 18/9.81

t = 1.83 [s], we found the answer for B.

Now using the following equation.

$y = y_{o} + v_{o}*t - 0.5*g*t^{2}\\$

where:

y = elevation [m]

Yo = initial elevation = 0

y = 18*(1.83) - 0.5*9.81*(1.83)²

y = 16.51 [m]

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Where:

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3 years ago

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