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

it has the same slope and a different y-intercept

hope this : )

15 batches are the most you can make

step-by-step explanation: