Question

James drops a water balloon from a height of 45 m. How long will it take the water balloon to hit the ground? Use the formula
`h(t)=-5.2+v_(0)t+h_(0)` where v0 is the initial velocity and h0 is the initial height. Round to the nearest tenth of a second.

Answer 1

When the water balloon hits the ground, its height is zero. Thus, we want to solve for
`h(t)=0`. Therefore,
`h(t)=0=-5.2t^(2)+ v_(0) * t +h_(0)`.
Since James drops the balloon, we know that its initial velocity is zero. Thus,
`v_(0)=0`.
Since James drops the balloon from a height of 45 m, its initial height is 45 m. Thus,
`h_(0)=45`.
Plugging these numbers into the equation, we get
`h(t)=0=-5.2t^(2)+(0)t+45=45-5.2t^(2)`.
Reducing this equation, we get
`5.2t^(2)=45`, then
`t^(2)=(225)/(26)`. Taking the square root of both sides, we get
`t=\sqrt{(225)/(16)}` and
`t=- \sqrt{(225)/(16)}`.
Since t can't be negative,
`t=\sqrt{(225)/(16)}=(15 √(26))/(26)`.