Janice tosses a softball directly upwards with an initial velocity of 100 feet per second. Use the formula h = vt - 16t^2 to answer each question.
a. When will the ball reach a height of 84 feet?
b. When will the ball hit the ground?
so, plug in the numbers.
solve for t when h=84 and 0
To find when the ball will reach a height of 84 feet, we can use the formula h = vt - 16t^2, where h is the height, v is the initial velocity, and t is the time.
a. The first step is to plug in the given values into the formula:
84 = 100t - 16t^2
b. Rearrange the equation to form a quadratic equation:
16t^2 - 100t + 84 = 0
c. To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. Let's use factoring in this case:
16t^2 - 100t + 84 = (4t - 6)(4t - 14) = 0
d. Set each factor equal to zero and solve for t:
4t - 6 = 0 or 4t - 14 = 0
4t = 6 or 4t = 14
t = 6/4 or t = 14/4
t = 1.5 seconds or t = 3.5 seconds
Therefore, the ball will reach a height of 84 feet at t = 1.5 seconds and t = 3.5 seconds.
To find when the ball will hit the ground, we need to find the time when the height (h) is equal to zero.
a. Plug in h = 0 into the formula:
0 = 100t - 16t^2
b. Rearrange the equation to form a quadratic equation:
16t^2 - 100t = 0
c. Factor out common terms:
t(16t - 100) = 0
d. Set each factor equal to zero and solve for t:
t = 0 or 16t - 100 = 0
t = 0 or 16t = 100
t = 0 or t = 100/16
t = 0 or t = 6.25 seconds
Therefore, the ball will hit the ground at t = 0 seconds and t = 6.25 seconds.
To answer these questions using the given formula h = vt - 16t^2, we need to substitute the values into the equation and solve for the unknown variable t.
a. When will the ball reach a height of 84 feet?
To find when the ball reaches a height of 84 feet (h = 84), we need to solve the equation:
84 = 100t - 16t^2
Rearranging the equation, we get:
16t^2 - 100t + 84 = 0
This is a quadratic equation, so we can solve it using factoring, completing the square, or the quadratic formula. To make the calculations easier, let's first try to simplify the equation by dividing every term by 4:
4t^2 - 25t + 21 = 0
Unfortunately, this equation cannot be factored easily, so we'll use the quadratic formula to solve it:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
Substituting the values into the quadratic formula, we get:
t = (-(-25) ± sqrt((-25)^2 - 4 * 4 * 21)) / (2 * 4)
t = (25 ± sqrt(625 - 336)) / 8
t = (25 ± sqrt(289)) / 8
t = (25 ± 17) / 8
Therefore, there are two possible values for t:
t1 = (25 + 17) / 8 = 42 / 8 = 5.25 seconds
t2 = (25 - 17) / 8 = 8 / 8 = 1 second
So, the ball will reach a height of 84 feet at 1 second and 5.25 seconds.
b. When will the ball hit the ground?
To find when the ball hits the ground, we need to find the time at which the height is zero (h = 0).
Substituting h = 0 into the equation, we get:
0 = 100t - 16t^2
Again, let's simplify the equation by dividing every term by 4:
0 = 25t - 4t^2
Now it is a quadratic equation in standard form. We can solve it by factoring, completing the square, or using the quadratic formula. To make it easier, let's rewrite it:
4t^2 - 25t = 0
Factoring out t from both terms:
t(4t - 25) = 0
Setting each factor equal to zero:
t = 0 or 4t - 25 = 0
From the second equation, we find:
4t = 25
t = 25/4 = 6.25 seconds
Therefore, the ball will hit the ground at 0 seconds (the initial toss) and 6.25 seconds.