Question

A Men Trhows A Ball Off The Top Of A Building That is 46 feet hight he trows the ball with an anitial velocity of 33 ft/sec

a. write the equation to model this problem
b. use the model to estimate the height of the ball at 2.5 seconds.
c. what is the ball's maximum height?

Answer 1

The height of the ball at 2.5 seconds is 42.75feet

How to model the equation of a body

The standard equation for the equation is modeled as:

$h(t)=t^2-v_0t+d$

Given the following parameters;

$v_0=33ft/sec\\d = 46ft$

Substitute into the expression to have:

$h(t)=t^2+33t-46$

The height of the ball at 2.5 seconds is xpressed as:

h(2.5) = (2.5)^2+33(2.5)-46
h(2.5) = 6.25+82.5-46

h(2.5) = 42.75 feet

Hence the height of the ball at 2.5 seconds is 42.75feet

The velocity of the ball at the maximum height is zero, hence;

v = dh/dt = 2t + 33

2t = 33

t = 16.5secs

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