Question

A body is projected with an initial speed of 100√3 ms^-1 at an angle of 60⁰ above the horizontal. If g = 10ms^-2 then velocity of the projectile
a) Is perpendicular to its acceleration at the instant t = 15 sec.
b) Is perpendicular to initial velocity of projection at t = 20 sec.
c) Is minimum at the highest point
d) Changes both in magnitude and direction, during its flight.
Mark the answer as

Moderate

Solution

Time of flight , $\large T = \frac{2usin\alpha}{g}$
$\fn_cm \large \therefore T=\frac {2x100 \sqrt3 xsin60^0}{10}sec=30sec$
Acceleration of the projectile is vertically downward.At the highest point on its trajectory , velocity of the projectile is horizontal.
The body arrives at the highest point on its trajectory at t = T/2 = 15 sec
Horizontal component of velocity remains unchanged and vertical component becomes zero at the highest point.
Therefore magnitude of velocity becomes minimum at the highest point.
Velocity becomes perpendicualr to the initial velocity at $\large t=\frac{u}{gsin\alpha}=\frac{100\sqrt3}{10sin60^0}sec=20sec$
Therefore , option -(1) is correct.

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