Is it calculus or is it just algebra?

Calculus Level 3

From a height of 64 feet, a ball is thrown vertically upward with a velocity of 24 feet per second. One second later, another ball is dropped from the same height. At what height above the ground do the two balls pass each other?

NOTE : Use only Calculus . Use also g = 32 f t / s 2 g=32 ft/s^2 .


The answer is 48.

Samuel Job Espartero by Samuel Job Espartero , 21, Philippines

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2 solutions

Mark Hennings
Oct 30, 2018

At time t t second after the second ball is released, the first ball is a height y 1 = 64 + 24 ( t + 1 ) − 1 2 g ( t + 1 ) 2 y_1 = 64 + 24(t+1) - \tfrac12g(t+1)^2 feet above the ground, while the second ball is a height y 2 = 64 − 1 2 g t 2 y_2 = 64 - \tfrac12gt^2 feet above the ground. Since y 1 − y 2 = 24 ( t + 1 ) − 1 2 g ( 2 t + 1 ) = ( 24 − g ) ( t + 1 ) + 1 2 g y_1 - y_2 = 24(t+1) - \tfrac12g(2t+1) = (24-g)(t+1) + \tfrac12g we see that the balls pass each other at time t = g 2 ( g − 24 ) − 1 = 48 − g 2 ( g − 24 ) t = \frac{g}{2(g-24)} - 1 = \frac{48-g}{2(g-24)} at which time the balls are a height 64 − 1 8 g ( 48 − g g − 24 ) 2 64 - \tfrac18g\left(\frac{48-g}{g-24}\right)^2 feet above the ground. Given a value of g g of 32 32 , the answer is 48 \boxed{48} feet.

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@Mark Hennings , we really liked your comment, and have converted it into a solution. If you subscribe to this solution, you will receive notifications about future comments.

Brilliant Mathematics Staff - 2 years, 7 months ago
Parth Sankhe
Oct 18, 2018

Function for y y coordinate of the first ball :

y 1 = 64 + ( 24 t − ½ g t 2 ) y_{1}=64+(24t-½gt^2) ; where g = 32 feet/sec

Function for y y coordinate of the second ball :

y 2 = 64 − ½ g ( t − 1 ) 2 y_{2}=64-½g(t-1)^2 ; as it was released a second later, (t-1) is used instead of t.

Solving these two equations gives us t = 2 t=2 and y = 48 y=48 .

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thanks!!! hope this helps! uhmmm...anyway, are there involved any calculus here in this part?

Samuel Job Espartero - 2 years, 7 months ago

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