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TOBY04 · May 17, 2012, 04:12 PM

A water balloon is launched horizontally with a speed of 5.2 m/s from a balcony that is 6.20 m above the ground. Determine the velocity of the balloon when it strikes the ground below.

ebaines · May 17, 2012, 04:20 PM

Hint: the horizontal component of velocity will remain a constant 5.2 m/s, and the vertical component of velocity when the object hits the ground will be the same as for an object that is dropped from a height of 6.2m. Can you complete the problem from here?

TOBY04 · May 17, 2012, 04:23 PM

nope, I thought deltady=1/2gt^2

ebaines · May 18, 2012, 05:47 AM

Originally Posted by TOBY04

nope, i thought deltady=1/2gt^2

The equation y=1/2 gt^2 is a good one to use to calculate the time it takes for the object to hit the ground. But to find the velocity you can use the formula:

$<br /> v_2^2 - v_1^2 = 2ad<br />$

For an object being dropped $v_1 = 0,\ a = g$ , and you know that $d = 6.2m$ . Solve for $v_2$ . That'll give you the vertical velocity of the object at impact. You already know the horizontal velocity - you can use Pythagoras to find the total velocity.

TOBY04 · May 18, 2012, 06:30 AM

Thanks