Use a(t)=-32 ft/second squared as the acceleration due to gravity. A ball is thrown vertically upward from the ground with an initial velocity of 56 feet per second. For how many seconds will the ball be going upward?

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To determine for how many seconds the ball will be going upward, we need to find the time it takes for the ball to reach its maximum height.

We can use the equation of motion for the vertical direction:

v(t) = v₀ + at

where:
v(t) is the velocity of the ball at time t,
v₀ is the initial velocity,
a is the acceleration,
t is the time.

Initially, the ball is thrown upward, so the initial velocity v₀ is positive 56 feet per second. The acceleration due to gravity, a, is negative 32 feet per second squared since it acts downward.

Therefore, we have:

v(t) = 56 - 32t

At the highest point of the ball's trajectory, its velocity will be zero. So we can set v(t) = 0 and solve for t:

0 = 56 - 32t

Rearranging the equation, we get:

32t = 56

Dividing both sides by 32:

t = 56/32

Simplifying the fraction, we have:

t = 7/4 = 1.75 seconds

Hence, the ball will be going upwards for 1.75 seconds.