The horizontal range of a projectile fired at an angle of 1515 is 50 m.50 m. If it is fired with the same speed at an angle of 45,45, its range will be:
1. 60 m60 m
2. 71 m71 m
3. 100 m100 m
4. 141 m141 m 

Hint: R=u2sin2θgR=u2sin2θg
 

Step: Find the range of a projectile.
The motion of the projectile is given by;

Given, θ=15θ=15 and R=50 mR=50 m
The range of a projectile is given by;
R=u2sin2θgR=u2sin2θg
Putting all the given values in the formula, we get;
R=50 m=u2sin(2×15)gR=50 m=u2sin(2×15)g
50×g=u2sin30=u2×1250×g=u2sin30=u2×12
50×g×2=u250×g×2=u2
u2=50×9.8×2=100×9.8=980u2=50×9.8×2=100×9.8=980
u=980u=980
u=49×20u=49×20
u=7×2×5 m/su=7×2×5 m/s
u=31.304 m/su=31.304 m/s
For θ=45,R=u2sin(2×45)g=u2g              (sin90=1)
R=(145)2g
R=14×14×59.8
R=100 m
Hence, option (3) is the correct answer.