MATH SOLVE

4 months ago

Q:
# HELP ME PLEASE I DONT UNDERSTAND THIS PROBLEM!!!!! Worth 20PTa ball is thrown with a slingshot at a velocity of 100 ft/sec at an angle 21 above the ground from a height of 5ft approximately how long does it take for the ball to hit the ground? acceleration due to gravity is 32 ft/s^2

Accepted Solution

A:

It takes approximately 6.3 seconds to hit the ground.

We use the equation h(t)=-1/2gt²+v₀t+h₀ to set this up. g is the acceleration due to gravity, v₀ is the initial velocity and h₀ is the initial height:

h(t) = -1/2(32)t² + 100t + 5

h(t) = -16t² + 100t + 5

We will use the quadratic formula to solve this:

[tex]t=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\=\frac{-100\pm\sqrt{100^2-4(-16)(5)}}{2(-16)} \\ \\=\frac{-100\pm\sqrt{10000--320}}{-32} \\ \\=\frac{-100\pm\sqrt{10320}}{-32} \\ \\=\frac{-100\pm101.59}{-32} \\ \\=\frac{-100+101.59}{-32}\text{ or }\frac{-100-101.59}{-32} \\ \\=\frac{1.59}{-32}\text{ or }\frac{-201.59}{-32} \\ \\=-0.05\text{ or }6.3[/tex]

Since a negative amount of time makes no sense, the answer is 6.3.

We use the equation h(t)=-1/2gt²+v₀t+h₀ to set this up. g is the acceleration due to gravity, v₀ is the initial velocity and h₀ is the initial height:

h(t) = -1/2(32)t² + 100t + 5

h(t) = -16t² + 100t + 5

We will use the quadratic formula to solve this:

[tex]t=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\=\frac{-100\pm\sqrt{100^2-4(-16)(5)}}{2(-16)} \\ \\=\frac{-100\pm\sqrt{10000--320}}{-32} \\ \\=\frac{-100\pm\sqrt{10320}}{-32} \\ \\=\frac{-100\pm101.59}{-32} \\ \\=\frac{-100+101.59}{-32}\text{ or }\frac{-100-101.59}{-32} \\ \\=\frac{1.59}{-32}\text{ or }\frac{-201.59}{-32} \\ \\=-0.05\text{ or }6.3[/tex]

Since a negative amount of time makes no sense, the answer is 6.3.