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A particle is projected with an initial velocity of 40 m/s in a direction making an angle 60 degree above horizontal. find the time after which the velocity of the particle is in a direction making an angle 45 degree above horizontal.

A particle is projected with an initial velocity of 40 m/s in a direction making an angle 60 degree above horizontal. find the time after which the velocity of the particle is in a direction making an angle 45 degree above horizontal.

Grade:11

2 Answers

Aman gohel
71 Points
6 years ago
(All velocities are in vertical Y-axis)Initial velocity(u)=40sin60°=34.64m/sFinal velocity(v)=40sin45°=28.28According to the equation of motion,V=u-gt=>28.28=34.64-(9.8*t)=>-6.36=9.8t=>t=0.648s(Ans)
Vikas TU
14149 Points
6 years ago
Dear Student,
the horizontal velocity in a projectile motion always remain constant
40cos60=ucos45
=>u=20 m/s is the velocity when the angle with horizontal is 45⁰.

Hence only the vertical component of the velocity changes

Veritcal initial velocity=u1=40sin60
vertical velocity when the hoz angle is 45⁰ =v1= 20 sin45

By equation of motion,
v=u+at
=>v1=u1-gt
=>20 -20=gt
=>t=1.5s
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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