Question

A ball is thrown directly downward with an initial speed of 7.05 m/s, from a height of 30.8 m. After what time interval does it strike the ground?

Answer 1

you would want to use the equation x(f)=x(i)+V(i)*t=(1/2)*a*t^2. set either your x(f) or x(i) equal to zero and the other to the height. you have your inital velocity and don't forget that the acceleration is gravity. plug in the numbers and then slove for t.

Answer 2

i tried that but i keep getting the wrong ans

Answer 3

I had this same problem and I used the equation that Stroodles77 showed. I plugged in the answer I got into my graphing calculator, graphed it, and then calculated the zeros. The answer was the zero on the negative x-axis.

Answer 4

Starting from \[s=v_{i}t+\frac{1}{2}at^2\] S=30.8m, v=7.05m/s. So, \[30.8m =7.08\frac{m}{s}t+\frac{1}{2}*9.8\frac{m}{s^2}t^2\] or \[0 =4.9\frac{m}{s^2}t^2+7.08\frac{m}{s}t -30.8m\] The quadratic equation gives us \[x=\frac{-7.08\pm \sqrt{7.08^{2}-4*4.9*(-30.8)}}{2*4.9}\]

Answer 5

Make that t =..., sorry.