Question:

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.

My attempt:

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

I did:

distance = rate*time

80m = rate*1.67 seconds

rate = 48m/s

then using t = 1.67 seconds, and v = 48m/s

Lower limit:

48sin(theta)(1.67) - 5(1.67)^2 = 1.2m

theta = 10.87 degrees

Upper limit:

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.