Mark throws a baseball straight up with an initial velocity of 60 ft / sec. Corey throws a baseball straight up with initial velocity 70ft/sec. If both the throws have an initial height 6 ft how much higher does Corey's throw reach than Mark's throw?

Corey's throw is higher than Mark's throw by 20.31 feet

Step-by-step explanation:

The vertical height of the object is given by : - 16t² + vt + d, where v is initial velocity and d is the initial height of the object.

Initial velocity of Mark = 60 ft/sec

Initial height of Mark's throw = 6 feet

Equation of vertical height : h (t) = - 16t² + 60 t + 6

Now, maximum height is achieved at time when h' (t) = 0

⇒ - 32t + 60 = 0

⇒ t = 1.875 seconds

Height of Mark's throw : h (1.875) = - 16 (1.875) ² + 60*1.875 + 6

⇒ Height of Mark's throw = 62.25 feet

Initial velocity of Corey = 70 ft/sec

Initial height of Corey's throw = 6 feet

Equation of vertical height : h (t) = - 16t² + 70 t + 6

Now, maximum height is achieved at time when h' (t) = 0

⇒ - 32t + 70 = 0

⇒ t = 2.1875 seconds

Height of Mark's throw : h (2.1875) = - 16 (2.1875) ² + 70*2.1875 + 6

⇒ Height of Corey's throw = 82.56 feet

Hence, Corey's throw is higher than Mark's throw by : 82.56 - 62.25

= 20.31 feet