A ball is thrown directly downward with an initial speed of 7.90 m/s, from a height of 30.2 m. After what time interval does it strike the ground?

Question

zephyr141 · Answer

kinematic equations. use one of the following.
$$d=v_{0}*t+\frac{1}{2}a*t^2$$$$v^2=v_0+2a*d$$$$v=v_0+a*t$$$$d=\frac{v_0+v}{2}*t$$you have initial velocity and height. assuming we aren't throwing balls off buildings on other planets our acceleration is going to be 9.8m/s^2 so now you're just missing time. which os these equations will help you find time with what information you have now?

zephyr141 · Answer

$$v^2=v_0^2+2*a*d$$ i messed up the second equation. forgot an exponent there.

zephyr141 · Answer

you'll have to use one to find the final velocity then using that plug it into another equation to find your t.

anonymous · Answer

you have to use        
s=ut+0.5att          . here u=7.9 and a= 9.8. after solving quadratic you will get the  time