A rock is launched into the air from the top of a 50 foot cliff at an initial velocity of 120 ft/sec What is the maximum height the rock will reach? (hint: Projectile motion formula is - 16t ^ 2 + vt + h

Step-by-step explanation:

The rock reaches its maximum height when the velocity is zero

Given the equation modeled by the height given as:

h(t) = - 16t^2 + vt + h

If v = 120ft/s and h = 50

h(t) = - 16t^2 + 120t + 50

First calculate t

At max height dh/dt = 0

dh/dt = -32t + 120

-32t +120 = 0

-32t = -120

t = -120/-32

t = 3.75s

Substitute t = 3.75 into the equation for the heightt

h(t) = - 16t^2 + 120t + 50

h(3.75) = - 16(3.75)^2 + 120(3.75) + 50

h(3.75) = - 225+ 450+ 50

h(3.75) = - 225+ 500

h(3.75) = 275m

Hence the max height reached by the rock is 275m

15

step-by-step explanation:

we are given the following expression and we are to solve it and find the product:

-13 × (-5)

solving the given expression in the standard order of solving a mathematical expression, beginning with the opening of brackets and then multiplying the terms and changing their signs.

-3 × -5 = 15 (answer is positive because when negative is multiplied with negative, it gives a positive answer)

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step-by-step explanation: