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Shivam Bhagat Grade: 9
        a boy throws a ball upwards with an angle of 45 degrees of velocity 100 m/s that makes a semicircle in the sky. what will be the max. height of the ball????
6 years ago

Answers : (5)

AKASH GOYAL AskiitiansExpert-IITD
419 Points
										

Dear Shivam


initial upward velocity u=100/√2 m/s


at highest point velocity is zero


using v2=u2-2gH


put v=0


H=u2/2g = 10000/(2x2x10)= 250m


 


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AKASH GOYAL


AskiitiansExpert-IITD


 


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6 years ago
vikas askiitian expert
510 Points
										

range of projectile = u2sin2@/g = u2/g             (at @=45)


it forms a semicircle in sky so this range is equal to diameter of this circle...


now , D = u2/g


D =2R                                          (R is radius of semicircle)


R = u2/2g


maximum height will be radius of this semicircle


Hmax = u2/2g=500m

6 years ago
Anirud Thyagharajan
14 Points
										

hello ,


since the formula for height of a projectile is u2sin2angle/2g, the answer can be calculated thus...

6 years ago
bhargav teja
33 Points
										

since it makes a semi circle in skymax height is half of i t's range 


  range R=U2SIN2(ANGLE)/g


     100*100/10


=1000


         then its max.height =500

6 years ago
Fawz Naim
37 Points
										

the ball will undergo a projectile motion. the formula for the max. height attained by the ball is =u^2cos^theta/2g


theta=45 deg


u=100m/s  g=10m/s^2


therefore


H=10000*0.5/2*10


=250m

6 years ago
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