# physics

posted by
**micole** on
.

A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 9.50 m/s at an angle of 24.0° below the horizontal. It strikes the ground 5.00 s later.

(a) How far horizontally from the base of the building does the ball strike the ground?

43.39 m

(b) Find the height from which the ball was thrown.

141.945 m

(c) How long does it take the ball to reach a point 10.0 m below the level of launching?

My answers for parts A and B are correct; I can't figure out C.

I tried using the equations delta y= Voy*t + (1/2)g*t^2, using 0 and 9.5sin24 as Voy, but neither worked. I solved the quadratic and got two irrational solutions or times greater than 5 seconds. My last attempt I used t=(2*distance/g)^(1/2)= 1.427seconds; this sounds somwhat logical. Could this be correct?

For (c), did you use -9.8 m/s^2 as the coefficient of t^2? . Did you use -10 m for y? You must consider the upward initial velocity component. Using the quadratic equation, I get a positive root of t=1.87 s.

I did try using both positive and negative g and 10. I also did one positive with one negative; I tried everything. This assignment is online, so I plugged in your answer and that's incorret also. I don't understand, but thanks for the help.