A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.5m/s at an angle of 17degrees below the horizontal. It strikes the ground 3 seconds later. how far horizontally from the base of the building does the ball strike the ground? What is the height from which the ball was thrown? How long does it take the ball to reach a point 10.0m below teh level of launching.
2006-09-30 07:40:33 · 2 answers · asked by Jayu P 1 in Science & Mathematics ➔ Physics
The ball strikes the ground 3*8.5*sin(17)=24.38 meters horizontally away from the house.
Next to determine the height from which the ball was thrown:
Initial downward velocity was sin(17)*8.5=2.48 m/s;
height=2.48*3+.5*9.8*9=51.55 meters.
Last question: 10=2.48*t+.5*9.8*t^2 Just use the quadratic equation to solve this to get:1.197 seconds.
2006-09-30 08:01:31 · answer #1 · answered by bruinfan 7 · 0⤊ 3⤋
use kinematic equations...
(8.5)cos(-17) = initial velocity in x direction
(8.5)sin(-17) = initial velocity in y direction
X(t) = X0 + Vt + at^2, so X(3) = (8.5)cos(-17)*(3)
you can do the rest... just use kinematic equation in Y direction, then for the 3rd question set Y(t) = -10
2006-09-30 07:59:14 · answer #2 · answered by fleisch 4 · 1⤊ 0⤋