KinematicsHard
Question
At what angle should a body be projected with a velocity 24 ms-1 just to pass over the obstacle 14 m high at a distance of 24 m. [Take g = 10 ms-2]
Options
A.tan θ = 19/5
B.tan θ = 1
C.tan θ = 3
D.tan θ = 2
Solution
x = 24 = u cosθ.t
⇒ t =
y = 14 = u sinθt −
gt2
⇒ 14 =
⇒ 14 = u tan θ − 5 sec2θ
⇒ 5 tan2θ − 24 tan θ + 19 = 0
⇒ tanθ = 1, 19/5. Ans
⇒ t =
y = 14 = u sinθt −
gt2⇒ 14 =
⇒ 14 = u tan θ − 5 sec2θ
⇒ 5 tan2θ − 24 tan θ + 19 = 0
⇒ tanθ = 1, 19/5. Ans
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