wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A projectile is given an initial velocity of (^i+2^j) m/s. Then, the equation of its trajectory is: [Take g=10 m/s2]


A
y=x5x2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
y=2x5x2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
4y=2x5x2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4y=2x25x2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B y=2x5x2
We know equation of trajectory of a projectile is
y=xtanθgx22u2cos2θ ...(1)
Here, initial velocity is u=^i+2^j


tanθ=YcomponentXcomponent
tanθ=21=2 ...(2)
Magnitude of initial velocity,
u=12+22=5 m/s...(3)
cosθ=basehypotenuse=15


On putting values in Eq. (1),
y=x×2gx22×(5)2×(15)2
y=2x10x22×5×15
y=2x5x2
Equation of its trajectory is y=2x5x2

flag
Suggest Corrections
thumbs-up
18
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon