integration problem (pls help)

Apr 2019
3
0
Philippines
If a ball is thrown vertically upwards from a platform 8 feet above the ground. After 2 seconds, the ball is 2feet below the platform. Find the initial velocity of the ball. (Acceleration due to gravity = -10m/s^2)

Answer key: 31m/s

how????
 

romsek

MHF Helper
Nov 2013
6,666
3,004
California
$s(t) = s_0 + v_0 t + \dfrac 1 2 g t^2$

$a \approx -10m/s^2$

$s_0 = 8ft$

$s(2) = 8-2=6ft$

Just plug everything in and you'll have some simple algebra to find $v_0$
 
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Apr 2019
3
0
Philippines
In still confused cause I keep ending up with 9m/s

6 = -5(2^2) +2Vo +8
9 = Vo
 
Last edited:
Dec 2014
131
102
USA
$v_0 = 9 \, m/s$ is the correct solution for the given parameters. If the initial velocity were 31 m/s, the projectile would take slightly more than 3 seconds to reach the top of its trajectory.
 

romsek

MHF Helper
Nov 2013
6,666
3,004
California
In still confused cause I keep ending up with 9m/s

6 = -5(2^2) +2Vo +8
9 = Vo
oh... it looks like there's some units conversion to be done.

They give the initial and final positions in feet but the acceleration due to gravity in m/s.

Convert the positions to meters.
 
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Apr 2019
3
0
Philippines
Ohhhh okay thanks so much. I finally got it!