Brian kicks a soccer ball with the initial velocity of 25m/sec. The equation h=vt-5t^2. At what time during the descent is the height of the soccer ball 20 meters?

Question

(where h=height, v= initial velocity, and t is the time.) represents this situation.

I also have this equation to go off of:

Y = - 5t^2 - 25t - 20

Answer 1

The problem seems to me to be ill-posed, because the initial angle of the trajectory is not given. The equation

y(t)=v0*t - (1/2)gt^2 is valid only if the given v0 pertains ONLY to the vertical component of the initial velocity of the ball after it's kicked. Different angles will, for a given initial speed, give different values of v0, and hence the quadratic equation solution upon which the answer is based will differ with angle.

Answer 2

The second equation for Y is irrelevant.

You want to know the time at which the height of the ball is 20.

In other words: which t value makes h = 20?

As an equation:

20 = vt - 5t²

v = 25 is given. So solve,

20 = 25t - 5t²

or in standard form:

5t² -25t + 20 = 0

You can use the quadratic formula(a=5,b=-25,c=20) but the quadratic on the left side of the equation will factor.