The horizontal range of a projectile fired at an angle of 15∘15∘ 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
Step: Find the range of a projectile.
The motion of the projectile is given by;
Given, θ=15∘θ=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∘)g⇒R=50 m=u2sin(2×15∘)g
⇒50×g=u2sin30∘=u2×12⇒50×g=u2sin30∘=u2×12
⇒50×g×2=u2⇒50×g×2=u2
⇒u2=50×9.8×2=100×9.8=980⇒u2=50×9.8×2=100×9.8=980
⇒u=√980⇒u=√980
⇒u=√49×20⇒u=√49×20
⇒u=7×2×√5 m/s⇒u=7×2×√5 m/s
⇒u=31.304 m/s⇒u=31.304 m/s
For θ=45∘,R=u2sin(2×45∘)g=u2g (∵sin90∘=1)
⇒R=(14√5)2g
⇒R=14×14×59.8
⇒R=100 m
Hence, option (3) is the correct answer.
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