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Q.

A Projectile travels at an angle θ below horizontal, 1sec after projection from origin as shown. What is the value of  θ ? (Take g= 10 m/s2)
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a

tan−1(12)

b

tan−1(2)

c

37°

d

53°

answer is B.

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Detailed Solution

tanθ=vyvx=uy−gtux          ux=41m/s           3=uy×1−12×10×12      uy=8 m/s                 ⇒  tan θ=|8−104|=|−12|

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A Projectile travels at an angle θ below horizontal, 1sec after projection from origin as shown. What is the value of  θ ? (Take g= 10 m/s2)