In Fig. 4-38, a ball is launched with a velocity of 8.50 m/s, at an angle of 50°
ID: 1691544 • Letter: I
Question
In Fig. 4-38, a ball is launched with a velocity of 8.50 m/s, at an angle of 50° to the horizontal. The launch point is at the base of a ramp of horizontal length d1 = 3.70 m and height d2 = 3.60 m. A plateau is located at the top of the ramp.(a) Does the ball land on the ramp or the plateau?
- ramp or plateau
(b) What is the magnitude of its displacement from the launch point when it lands?
--------m
(c) What is the angle of its displacement from the launch point when it lands?
(counterclockwise from the horizontal axis directed to the right)
Explanation / Answer
Vox=vo*cos(50 degrees)=5.464 m/s Voy=vo*sin(50 degrees) =6.5114 m/s max ht : voy^2/(2*g) =2.1632 meters max ht time : voy/g =.66443 s a) the ball will land on the ramp b) the ball bounces off the ramp c) look at b, can not determine total flight time is max ht t * 2 = 1.329 s max distance : vox*t =7.26 meters firing from 3.7 meters to a ramp of 3.6 meters in height, the ball will hit the wall of the ramp and bounce off because the height in not high enough for the ball to reach the top of the ramp vo must be : vo=((9.8 m/s^2*(3.7 m)^2)/(2*cos(50 degrees)^2*(3.7m*tan(50 degrees)-3.6m)))^1/2