-8 = 0 + 20 t - 4.9 t^2
4.9 t^2 -20 t -8 = 0
solve quadratic for t
v = 20 - 9.8 t
A rock is tossed straight up with an initial speed of 20 m/s. When it returns, it fall into a hole that is 8.0 m deep. What is the velocity of the rock (in m/s) when it hits the bottom of the hole? Use -9.80 m/s2 for the local acceleration due to gravity and assume no significant air resistance. Assume that the y axis is positive in the up direction
3 answers
I am sorry but I am still confused...
more details:
h = original height + Vi t - (1/2)(9.8) t^2
-8 = 0 + 20 t - 4.9 t^2
4.9 t^2 -20 t -8 = 0
solve quadratic for t
velocity = original velocity - 9.8 t
v = 20 - 9.8 t
h = original height + Vi t - (1/2)(9.8) t^2
-8 = 0 + 20 t - 4.9 t^2
4.9 t^2 -20 t -8 = 0
solve quadratic for t
velocity = original velocity - 9.8 t
v = 20 - 9.8 t