 # 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???? AKASH GOYAL AskiitiansExpert-IITD
419 Points
11 years ago

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

All the best.

AKASH GOYAL

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11 years ago

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

11 years ago

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

11 years ago

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