Question

Mitch throws a baseball straight up in the air from a cliff that is 95 ft high. The initial velocity is 95 /ftsec. The height (in feet) of the object after t sec is given by =ht+−19t2+95t95. Find the time at which the height of the object is 114 ft. Round your answers to two decimal places.

Answer 1

The time at which the height of the object is 114 ft rounded to two decimal places is; 5.85 seconds

How to find the height of a projectile?

We are given;

Height of cliff = 95 ft

Initial velocity of throw = 95 ft/s

The height (in feet) of the object after t sec is given by;

h(t) = −19t² + 95t + 95.

Now, we want to find the time at which the height of the object is 114 ft. Thus, we will set h(t) = 114 ft to get;

114 = −19t² + 95t + 95.

Subtract 114 from both sides to get;

−19t² + 95t + 95 = 0

Using quadratic formula, we have;

t = [-95 ± √(95² - 4(-19 * 95))]/(2 * -19)

t = [-95 ± √(16245)]/-38

t = (-95 ± 127.4559)/(-38)

t = (-95 - 127.4559)/-38 or (-95 + 127.4559)/-38

t = 5.85 s or -0.85

Time cannot be negative and so t = 5.85 seconds

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