The formula for height, in feet of a projectile under the influence of gravity is given by h=-16t^2+vt+s, where t is the time in seconds, v is the upward velocity at the start, and s is the starting height. Marvin throws a baseball straight up into the air at 70 feet per second. The ball leaves his hand at a height of 5 feet. When does the ball reach a height of 75 feet?

1.55s

Step-by-step explanation:

The height equation of the projectile is given as;

h = -16t^2 + vt + s 1

where;

t is the time in seconds,

v is the upward velocity at the start,

s is the starting height.

Given;

v = 70 ft/s

s = 5 ft

Substituting the values equation 1 becomes;

h = -16t^2 + 70t + 5

For the ball to reach 75 ft, h = 0

75 = -16t^2 + 70t + 5

-16t^2 + 70t + 5 - 75 = 0

-16t^2 + 70t - 70 = 0

solving the quadratic equation;

Time t = 2.83 s and t = 1.55s

Therefore, the ball will first reach 75 ft after 1.55 seconds.

51-50

step-by-step explanation:

this answer is !