MATH SOLVE

6 months ago

Q:
# 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 [tex]h(t)=-5.2+v_{0}t+h_{0}[/tex] where v0 is the initial velocity and h0 is the initial height. Round to the nearest tenth of a second.

Accepted Solution

A:

When the water balloon hits the ground, its height is zero. Thus, we want to solve for [tex]h(t)=0[/tex]. Therefore, [tex]h(t)=0=-5.2t^{2}+ v_{0} * t +h_{0}[/tex].

Since James drops the balloon, we know that its initial velocity is zero. Thus, [tex]v_{0}=0[/tex].

Since James drops the balloon from a height of 45 m, its initial height is 45 m. Thus, [tex]h_{0}=45[/tex].

Plugging these numbers into the equation, we get [tex]h(t)=0=-5.2t^{2}+(0)t+45=45-5.2t^{2}[/tex].

Reducing this equation, we get [tex]5.2t^{2}=45[/tex], then [tex]t^{2}=\frac{225}{26}[/tex]. Taking the square root of both sides, we get [tex]t=\sqrt{\frac{225}{16}}[/tex] and [tex]t=- \sqrt{\frac{225}{16}}[/tex].

Since t can't be negative, [tex]t=\sqrt{\frac{225}{16}}=\frac{15 \sqrt{26}}{26}[/tex].

Since James drops the balloon, we know that its initial velocity is zero. Thus, [tex]v_{0}=0[/tex].

Since James drops the balloon from a height of 45 m, its initial height is 45 m. Thus, [tex]h_{0}=45[/tex].

Plugging these numbers into the equation, we get [tex]h(t)=0=-5.2t^{2}+(0)t+45=45-5.2t^{2}[/tex].

Reducing this equation, we get [tex]5.2t^{2}=45[/tex], then [tex]t^{2}=\frac{225}{26}[/tex]. Taking the square root of both sides, we get [tex]t=\sqrt{\frac{225}{16}}[/tex] and [tex]t=- \sqrt{\frac{225}{16}}[/tex].

Since t can't be negative, [tex]t=\sqrt{\frac{225}{16}}=\frac{15 \sqrt{26}}{26}[/tex].