Questions
The horizontal range of a projectile fired at an angle of 150 is 50 m. If it is fired with the same speed at
an angle of 45°, its range will be
detailed solution
Correct option is C
Given, θ = 15o and R = 50mRange, R=u2sin2θgPutting all the given values in the formula, we getR=50 m=u2sin2×15°9⇒50×g=u2sin30°=u2×12⇒50×g×2=u2⇒ u2=50×9.8×2=980⇒ u=980 m/s =145 m/s For θ=45°,R=u2sin2×45°g=u2g ∵sin90°=1⇒ R'=(145)29=14×14×59.8=100 mTalk to our academic expert!
Similar Questions
A ball is thrown with a velocity of u making an angle with the horizontal. Its velocity vector normal to initial vector (u) after a time interval of
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
