A ball is thrown vertically upward from a height of 2 m with initial velocity of 20 m/s (how high with the ball go? assume the acceleration is a(t)= -9.8 m/s)

Answer 1

Pls helppppp

Answer 2

h(t) = -4.9t^2 + 20t + 2
as with all quadratics, the vertex is at t = -b/2a = 20/9.8
so find h there.

Answer 3

But where did you get the 4.9??

Answer 4

it's 1.2 g, which they gave you

h(t) = 1/2 gt^2 + v0*t + h0

better review the basic formulas of ballistic trajectories.

Answer 5

In this question we start with the equation for the acceleration of the ball, which is -32 feet per second. Then, we take the antiderivative of this function in order to find our equation for velocity, which is -32t + C. In order to find out what C is, we use our initial velocity, which is 40. Thus, our equation for velocity is -32t + 40. Then, we need to figure out what the time is when the height of the ball is 0. Thus, we must take the antiderivative of the velocity equation, which results in -16t^2 + 40t + C. Again, in order to figure out what C is, we simply use the initial height of 500. This gives us a positional equation of -16t^2 + 40t + 500. Then, in order to figure out at what time the ball hits the ground, we set the positional equation to 0. This gives us the equation 0 = 16t^2 + 40t + 500. Then, once we factor this using the quadratic formula, we get (5/4) +/- (5*sqrt(21))/4, or t = −4.47822 and t = 6.97822. Now, as time cannot be negative, then t must equal 6.97822. Then we simply put this value into our velocity equation, and get -183.303 as our final answer. (rounded)