1. A body is projected with an initial velocity of 45 m/s and reaches its max. height at 4 seconds. Find the angle of projection.
Answer: 60 degrees, 35 minutes.
My working:
u= 45sinA, a=-9.8, t=4.
v^2=u^2+2as
0 = [45sinA]^2 - 19.6*h
s=ut+0.5*at^2
h=180sinA-4.9(16)
h=180sinA-78.4
0= [45sinA]^2 - 19.6*[180sinA-78.4]
2025.sin^2(A)-3528+1536.64=0
Then i just used the quadratic formula, but didn't get the correct answer.
Answer: 60 degrees, 35 minutes.
My working:
u= 45sinA, a=-9.8, t=4.
v^2=u^2+2as
0 = [45sinA]^2 - 19.6*h
s=ut+0.5*at^2
h=180sinA-4.9(16)
h=180sinA-78.4
0= [45sinA]^2 - 19.6*[180sinA-78.4]
2025.sin^2(A)-3528+1536.64=0
Then i just used the quadratic formula, but didn't get the correct answer.
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