Question

a baseball player bats a ball straight up with an initial speed of 109 miles. it reaches a height of s = 0.03t-0.003t ^ 2 miles after t seconds. What is the speed of the ball when it is .0830 miles from the ground?

Answer 1

Solution-

Use initial velocity and gravity to find the equaiton for the height(S) of the ball,

v = 109 miles/hr = 0.030 miles/s

g = 9.81 m/s^2 = 6.21X10-4 miles/s2

also, we can use the third equation of motion to find the velocity at a certain height,

u=0.03 miles/s

g = 6.21X10-4 miles/s2

s=0.0830 miles

or

Therefore, at height, 0.830 miles ball speed will be 101.63 miles/hr.

NOTE:

The equation for height is given as,

S in miles and t in seconds.

First, find the value of time when the ball reaches a height of 0.0830 miles,

solve for t,

Imaginary roots (error in the equation)