Question

The record distance in the sport of throwing cowpats is 81.1 m. This record toss was set by Steve Urner of the United States in 1981. Assuming the initial launch angle was 45° and neglecting air resistance, answer the following.

(a) Determine the initial speed of the projectile.

Since only the x position and launch angle are given, I cannot figure out how to solve for initial velocity with the equations I have from the text.

Answer 1

Now the record distance is R=81.1 m Let the initial velocity of the projectile be vo The angle of projection given is 0-45 Then this record distance is given as R.= (vo cos θ)T 7: Time of flight of the projectiles 2v, sin θ Then R-(Vo cosins 26 Here 0 45 Then 81. 1mYo'sin 90* Then vo- 28.19mls Now time of flight is 7-2 = As the angle of projecti on increases beyond 45 then initial velocity vo increases upto the angle of projection 90 and then decreases. 9.8ml 228 1ml o)in 45 4067 v sin 9.8mls

Answer 2

R = 2 V^2 sinø cosø /g

V = v R . g / 2 sin45 . cos 45 = v 81.1 . 9.8 = 28.2 m/s

b) t = d / v = 81.1 / 28.2 . cos 45 = 4.068 = 4.07 s

c) The range would get less. The starting speed would be about the same. The time in the air would increase but with less horizontal speed the range would decrease