Question

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) 18.0 ❝ , (b) 32.0 ❝ , (c) 38.0 ❝ ?

Answer 1

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)
? = 21
4.013 t-4.9t^2+1.524,
t =1.1 s
R =11.2*cos 21*1.1 = 11.5m

b)
? = 34
6.26t-4.9t^2+1.524,
t = 1.49s
R =11.2*cos 34*1.49 = 13.83.m

c)
? = 44
7.78 t-4.9t^2+1.524,
t = 1.76s
R = 11.2*cos 44*1.76 = 14.18.m
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TO AVOID SOLVING QUADRATIC EQUATION

Using v^2- u^2 = 2as, find the final speed.

a)
v = ? [(11.2 sin 21) ^2 + 19.6*1.524]
v = ? [(11.2 sin 21) ^2 + 29.87]
v = - 6.78m/s minus since it is falling down
Time t = distance / average velocity = - 1.524*2 / (11.2 sin 21 -6.78)
t = 1.1 s
R =11.2*cos 21*1.1 = 11.5m

b)
v = ? [(11.2 sin 34) ^2 + 29.87]
v = - 8.31 m/s minus since it is falling down
t = - 1.524*2 / (11.2 sin 34- 8.31)
t = 1.49 s
R =11.2*cos 34*1.49 = 13.83.m

c)
v = ? [(11.2 sin 44) ^2 + 29.87]
v = - 9.51 m/s minus since it is falling down
t = - 1.524*2 / (11.2 sin 44- 9.51)
t = 1.76 s
R = 11.2*cos 44*1.76 = 14.18.m