Douglas B. answered • 05/08/20
Algebra tutor with masters degree in applied math
Every projectile motion problem, where the projectile is close to the surface of the earth, is parabolic. That is, the equation for the height (h) of the projectile is
h(t) = h0+v0sin(θ)t+(1/2)gt^2, where t is time, h0 is initial height above ground, v0 is initial velocity, θ is the angle the projectile is shot above horizontal (For example, if the projectile is shot straight up, θ=90º), and g is acceleration due to gravity (e.g. -9.8 m/s^2).
So, the time when projectile hits the ground is the time for which h = 0. That is, you need to solve the quadratic equation
0 = h0+v0sin(θ)t+(1/2)gt^2 for t. Make sense?
Douglas B.
Do you know the quadratic formula? Regardless, I think you are on the right track. The vertex will allow you to find the max height (H) of projectile. From here you can just solve H = 0.5*g*t^2 for t. This will be the time it hits the ground. The only other option I can think of would be to try to factor the quadratic equation.05/08/20
Marina M.
So would the answer be that you need to A Find the vertex B Set the equation equal to zero C Set the equation equal to c D Divide the equation by 205/08/20