A shot-putter throws the shot with an initial speed of 11.2 m/s from a height of 5.00 ft above the ground. What is the range of the shot if the launch angle is

(a) 25.0 degrees ,

(b) 33.0 degrees ,

(c) 39.0 degrees

Answer:

5.00 ft = 1.524 m
In all the cases find the time to fall vertically through a distance of - 1.524 m

11.2 sin ?*t-4.9t^2+1.524 = 0
Range is given by
R = u cos? *t


a)
? = 25
4.733 t-4.9t^2+1.524,
t =1.22 s
R =11.2*cos 25*1.22 = 12.38m

b)
? = 33
6.099t-4.9t^2+1.524,
t = 1.458s
R =11.2*cos 33*1.49 = 13.695.m

c)
? = 39
7.048 t-4.9t^2+1.524,
t = 1.629s
R = 11.2*cos 39*1.76 = 14.1788.m