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`        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????`
8 years ago 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

All the best.
AKASH GOYAL

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```
8 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
```
8 years ago
```							hello ,
since the formula for height of a projectile is u2sin2angle/2g, the answer can be calculated thus...
```
8 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
```
8 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
```
8 years ago
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