Kinematics 5 Question 1

3. The trajectory of a projectile near the surface of the earth is given as y=2x9x2.

If it were launched at an angle θ0 with speed v0, then (Take, g=10ms2 )

(2019 Main, 12 April I)

(a) θ0=sin115 and v0=53ms1

(b) θ0=cos125 and v0=35ms1

(c) θ0=cos115 and v0=53ms1

(d) θ0=sin125 and v0=35ms1

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Answer:

Correct Answer: 3. (c)

Solution:

  1. Given, g=10m/s2

Equation of trajectory of the projectile,

y=2x9x2

In projectile motion, equation of trajectory is given by

y=xtanθ0gx22v02cos2θ0

By comparison of Eqs. (i) and (ii), we get

tanθ0=2 and g2v02cos2θ0=9 or v02=g9×2cos2θ0

From Eq. (iii), we can get value of cosθ and sinθ

cosθ0=15 and sinθ0=25

Using value of cosθ0 from Eq. (v) to Eq. (iv), we get

v02=10×(5)22×(1)2×9=10×52×9v02=259 or v0=53m/s

From Eq. (v), we get

θ0=cos115



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