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If a ball is thrown vertically upward with an initial velocity of 80 ft/sec, what would be the height of the ball after 2 seconds?

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Final answer:

The ball thrown vertically upward with an initial velocity of 80 ft/sec will reach a height of 96 feet after 2 seconds, calculated using the kinematic equation for uniformly accelerated motion under gravity.

Step-by-step explanation:

When a ball is thrown vertically upward with an initial velocity, its height after a certain time can be calculated using the kinematic equation for uniformly accelerated motion (assuming the only acceleration is due to gravity). The formula is:

h = v0t - ½gt2

For this problem, the initial velocity v0 is 80 ft/sec, acceleration due to gravity g is 32 ft/sec2 (downwards), and the time t is 2 seconds.

Substituting these values into the equation:

h = (80 ft/sec)(2 sec) - ½(32 ft/sec2)(2 sec)2

h = 160 ft - 64 ft

h = 96 ft

Therefore, the height of the ball after 2 seconds would be 96 feet.

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