An archer hits a target 80m away and the time taken to do this was 1.67 seconds. The target has a radius of 60cm, and its stands at a height of 1.8m, with the 1.8m mark being its centre. What range of angles could have been used for the archer to hit anywhere on the target.
v(t) = vcos(theta) + ( vsin(theta) - 10t) [gravity = 10m/s/s]
s(t) = vcos(theta) *t + ( vsin(theta) *t - 5t^2)
Distance = 80m, Time = 1.67 seconds
Upper height of target = 1.8m - 0.6m = 1.2m
Lower height of target = 1.8m + 0.6m = 2.4m
distance = rate*time
80m = rate*1.67 seconds
rate = 48m/s
then using t = 1.67 seconds, and v = 48m/s
48sin(theta)(1.67) - 5(1.67)^2 = 1.2m
theta = 10.87 degrees
48sin(theta)(1.67) - 5(1.67)^2 = 2.4m
theta = 11.79 degrees
I am not sure if this is the right way to do this question.