Mechanics> A stone is launched upward at 45 degrees ...

A stone is launched upward at 45 degrees with speed v₀.A bee follows the trajectory of the stone at a constant speed equal to the initial speed of the stone .
a)find the radius of curvature at the top point of the trajectory.
b)what is the acceleration of the bee at the top point of the trajectory?For the stone ,neglect the ir resistance

rahul , 10 Years ago

Grade 11

2 Answers

Sumit Majumdar

Last Activity: 10 Years ago

Dear student,
The path of the stone is going to be a parabolic path.
The maximum height reached by the stone would be given by:
$H=\frac{v_{0}^{2}sin^{2}\theta }{2g}$

a) Since the bee is following the path of the stone, so the path traversed by the stone would be parabolic as well and hence the radius of curvature would be given by:
$R=\frac{v_{0}^{2}sin^{2}\theta }{2g}=\frac{v_{0}^{2}sin^{2}\left (45^{0} \right ) }{2g}=\frac{v_{0}^{2}}{4g}$
b) The centripetal acceleration of the bee would be given by:
$a_{cent}=\frac{v_{0}^{2}}{R}=4g$
Regards
Sumit

Alpha

Last Activity: 7 Years ago

Centripetal acceleration = v^2/RA) velocity at top most point of the particle = u sin(45°) = u/√2Acceleration perpendicular to velocity of particle is `g` which is equal to centripetal acceleration at that moment for the particle`s journey. Let R be the the radius of curvature.=> g=((u/√2)^2)/R=> R=(u^2)/2gB) For bee the radius of curvature of the path will be same i.e. R=(u^2)/2gWe know that the speed of the bee is u at the highest point also. Let the centripetal acceleration at the highest point of the trajectory for bee be `a`.=> a=(u^2)/((u^2)/2g)=> a=2g