A ball is thrown from an initial height of 2 meters with an initial upward velocity of 25m/s. The ball's height h (in meters) after t seconds is given by the following.

Find all values of t for which the ball's height is 12 meters.
Round your answer(s) to the nearest hundredth.

To find when the ball's height is

7
meters, we substitute
7
for
h
and solve for
t
.
=7+2−25t5t2
In order to solve for
t
, we first rewrite the equation in the form
=+at2+btc0
.
=+−5t225t50
Next, we use the quadratic formula to solve for
t
.
=t−b±−b24ac2a
Our equation has
=a5
,
=b−25
, and
=c5
.
Another way
We use these values in the formula.
t
=−−25±−−252·4·55·25
=25±52510
We get that
t
can be either of two values to solve the equation.
=t=−25525100.2087…
or
=t=+25525104.7912…
Rounding these values to the nearest hundredth, we get
=t0.21
or
=t4.79
.
So, the ball's height is
7
meters at approximately
0.21
seconds (on its way up) or
4.79
seconds (on its way down).

To find the values of t for which the ball's height is 12 meters, we can use the equation for the ball's height h in terms of time t:

h = -4.9t^2 + v0t + h0

where h is the height, t is the time, v0 is the initial velocity, and h0 is the initial height.

In this case, h0 = 2 meters and v0 = 25 m/s, so the equation becomes:

h = -4.9t^2 + 25t + 2

We want to find the values of t for which h = 12. So we can set up the equation:

12 = -4.9t^2 + 25t + 2

Rearranging this equation:

4.9t^2 - 25t + 10 = 0

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 4.9, b = -25, and c = 10. Plugging these values into the quadratic formula, we get:

t = (-(-25) ± √((-25)^2 - 4(4.9)(10))) / (2(4.9))

Simplifying this equation:

t = (25 ± √(625 - 196)) / 9.8

t = (25 ± √429) / 9.8

Using a calculator, we can find the two approximate values of t:

t ≈ 0.55 and t ≈ 4.96

Therefore, the values of t for which the ball's height is 12 meters are approximately 0.55 seconds and 4.96 seconds.

To find the values of t for which the ball's height is 12 meters, we can use the given equation for the height of the ball:

h(t) = -4.9t^2 + v0t + h0

where h(t) represents the height of the ball at time t, v0 represents the initial upward velocity (25 m/s in this case), and h0 represents the initial height (2 meters in this case).

Substituting the given values into the equation, we have:

12 = -4.9t^2 + 25t + 2

Now, we need to solve this quadratic equation for t. We can rearrange the equation to bring all terms to one side:

4.9t^2 - 25t + 10 = 0

Since the equation is in the form of a quadratic, we can use the quadratic formula to solve for t:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 4.9, b = -25, and c = 10. Substituting these values into the quadratic formula, we get:

t = (-(-25) ± √((-25)^2 - 4(4.9)(10))) / (2 * 4.9)

Simplifying further, we have:

t = (25 ± √(625 - 196)) / 9.8
t = (25 ± √429) / 9.8

Calculating the values using a calculator, we get two possible solutions for t:

t ≈ 4.55 seconds
t ≈ 0.45 seconds

Therefore, the ball's height is 12 meters at approximately t = 4.55 seconds and t = 0.45 seconds.

h(t) = 2+25t-4.9t^2

h=12 when

2+25t-4.9t^2 = 12
t= 0.44 and 4.66 seconds