# A naughty student throws a water balloon straight down with a speed of 5 ft/s from a window 40 ft above the ground. (a) When will the balloon hit the head of an innocent 6 ft tall passerby? (b)...

A naughty student throws a water balloon straight down with a speed of 5 ft/s from a window 40 ft above the ground.

(a) When will the balloon hit the head of an innocent 6 ft tall passerby?

(b) What will be its speed when it hits?

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### 1 Answer

Hello!

We'll ignore air resistance.

The height of the balloon is

`H(t)=H_0-V_0*t-(g*t^2)/2,`

where `H_0=40` ft is the initial height, `V_0=5` ft/s is the initial downward velocity and `g=32ft/s^2` is the gravity acceleration.

The (downward) speed of the balloon is

`V(t)=V_0+g*t.`

Let's find time `t_1` for which `H(t_1)=6` ft (the height of a passerby):

`40-5t_1-16t_1^2=6,`

or

`16t_1^2+5t_1-34=0.`

This is a quadratic equation for `t_1,` and `t_1` must be >0, so

`t_1 = (-5+sqrt(25+4*16*34))/32 = (-5+sqrt(2201))/32 approx 1.31 (s).`

Speed will be 5+16*1.3=25.8 (ft/s).

The answers: **(a) 1.3 s (b) 25.8 ft/s**.