Two object, A and B vertical thrown up with velocities, 8ms^-2 and 100ms^-2 at 2 sec intervals. Where and when will the two objects meet each other. (G = 10ms^-2)
Answers
Answer:
Aperture of a lens is the effective diameter of its light transmitting area. Therefore, the brightness i.e. the intensity of the image formed by a lens which depends on the amount of light passing through the lens will vary as the square of the aperture of the lens.
Bhus soorrrrrryyyyyyyyyyy
Explanation:
Answer:
A: A time to collision is needed, given initial velocities and gravitational acceleration. That looks like a natural question for one of the four basic Kinematic Equations, namely Distance = v*t + 1/2 *a*t^2
Call them ball 1 and ball 2:
Ball 1:
distance = 50 * t + 1/2* 10 * t^2 = 50*t - 5*t^2 (taking g as approx 10 m/s^2)
Ball 2: distance = 40 * (t-2) - 1/2 * 10* (t-2)^2
We want the moment when their distance above the throw point is the same, so we equate the equations:
50*t -5*t^2 = 40*t - 80 - 5* (t^2 -4*t +4) = 40*t -80 -5*t^2 +20*t - 20
Hence: 10*t = 100 and t = 10 seconds after the first object launched.
Notice the question did NOT ask the distance from the throw point at which the objects meet - that turns out to be zero meters - the start point!