A ball is dropped from a height of 784 ft. It's height h, in feet, after t seconds is given by h(t)=784-16t^2. After how long will the ball reach the ground?

3 Answers
Apr 15, 2018

#t= 7 sec#

Explanation:

So here height is given as function of time, and I want to know when the height is equal to 0, so set it equal to 0:

#0= -16t^2+784#

The goal here is to solve for #t#:

Subtract #784# from both sides
#-16t^2=-784#

Divide by #-16#:
#t^2= 49#

Square root both sides:
#t =+-7 #

Since time can't be negative:
#t= 7 sec#

graph{-16x^2+784 [-10, 10, -5, 5]}

Apr 15, 2018

#t=7#

Explanation:

Since you are trying to find when the ball hits the ground, #h(t)=0# because the ground is at #0# feet.
#h(t)=0=784-16t^2# Isolate #t# and its accompanying exponents/coefficients.
#16t^2=784# Divide by 16 to isolate #t# and its exponent.
#t^2=49# Find the square root of each side.
#t=sqrt(49)# #(7^2=49)#
#t=+ or - 7# Since #t# represents time, it must be positive, so
#t=7#

Apr 16, 2018

#7# seconds

Explanation:

The height of the ball is given by the function #h(t)=784-16t^2#.

When the ball is on the ground, we can say that its height is #0#.

And so,

#784-16t^2=0#

#16t^2=784#

#t^2=784/16=49#

#t=sqrt(49)#

#t=+-7#

But since #t# is the time, and time cannot be negative, we only take the positive root, that is:

#t=7#

So, it'll take seven seconds for the ball to hit the ground.