A ball is thrown vertically upwards from a height of 6 ft with an initial velocity of 60 ft per second. How high will the ball go? Note that the acceleration of the ball is given by a(t)= 32 feet per second per second.

Well, if the ball is thrown vertically upwards with an initial velocity of 60 ft/s, it will eventually reach a point where its velocity becomes zero before it starts coming back down. This point is known as the highest point of the ball's trajectory.

To find the height the ball will go, we need to determine the time it takes for the ball to reach its highest point. We can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

At the highest point, the final velocity of the ball is zero. Therefore, we can rewrite the formula as 0 = 60 - 32t.

Solving this equation, we find t = 60/32 = 1.875 seconds.

Now, to find the maximum height reached by the ball, we can use the formula h = u*t - 0.5*a*t^2, where h is the height, u is the initial velocity, a is the acceleration, and t is the time.

Plugging in the values, we get h = 60*1.875 - 0.5*32*(1.875)^2.

Doing the math, we find that the ball will go approximately 67.5 feet high.

So, the answer is 67.5 feet. But hey, just imagine how thrilled the ball must be reaching such great heights! It's like becoming the superstar of gravity-defying tricks!

To find out how high the ball will go, we need to calculate the maximum height reached by the ball during its vertical motion.

We can use the kinematic equation:

v^2 = u^2 + 2as

where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement

Given:
u = 60 ft/s (initial velocity)
a = -32 ft/s^2 (acceleration due to gravity, downward direction)

At its maximum height, the final velocity, v, will be zero. Therefore:

0 = (60 ft/s)^2 + 2(-32 ft/s^2)(s)

Expanding this equation, we get:

0 = 3600 ft^2/s^2 - 64 ft/s^2 x s

Rearranging the equation, we have:

64 ft/s^2 x s = 3600 ft^2/s^2

Dividing both sides by 64 ft/s^2, we get:

s = 3600 ft^2/s^2 ÷ 64 ft/s^2

s = 56.25 ft

Therefore, the ball will reach a maximum height of 56.25 ft.

To determine the height the ball will reach, we can use the kinematic equation of motion for vertical motion:

h(t) = h0 + v0 * t - (1/2) * g * t^2

where:
h(t) is the height of the ball at time t
h0 is the initial height (6 ft)
v0 is the initial velocity (60 ft/s)
g is the acceleration due to gravity (32 ft/s^2)
t is the time

In this case, we want to find the maximum height which occurs when the ball reaches its peak and its velocity becomes zero. At this point, we can set v(t) = 0, and solve for time t.

0 = v0 - g * t
t = v0 / g

Substituting this value of t back into the equation for height, we can calculate the maximum height the ball will reach.

h_max = h0 + v0 * (v0 / g) - (1/2) * g * (v0 / g)^2

Now let's substitute the given values into the equation:

h_max = 6 + 60 * (60 / 32) - (1/2) * 32 * (60 / 32)^2

Simplifying the expression:

h_max = 6 + 112.5 - 30

h_max = 88.5 ft

Therefore, the ball will reach a maximum height of 88.5 ft.

h(t) = 6+60t-16t^2

max height at t = -60/-32 = 15/8
h(15/8) = 62.25 ft