Question

91. A particle is projected from the ground
at an angle 60° with horizontal with a
speed 72 km/hr. Taking g=10 m/s², the
time after which the speed of the
particle becomes half of its initial speed
is seconds
(1)
$\sqrt{2}$

(2)
$\sqrt{3}$

(3) √5
(4) 4​

Answer 1

SOLUTION.

Here we can use the knowledge of our vector addition. and I solve the question by splitting the question in two direction one is along X and other is along y.

Let after time t the velocity become half to that of initial.

Convert all the data in meters and seconds.

In Y direction.

Initial velocity = 20Sin60° m/sec

Final velocity after time t = v

Acceleration = 10m/sec²

ACC to EQ of motion

v = u - gt

v = 10√3 -10t.....(1)

In X direction.

Initial velocity = 20 Cos60°m/sec

Final velocity after time t remain the same because the Acceleration in X direction is zero.

Final velocity = 10m/sec.....(2)

As these two velocity are perpendicular to each other than the resultant can be written as ...

√(1)²+(2)²

V = √(10√3 - 10t)²+(10)²

ACC to question

V = 20/2

V = 10m/sec

10 = √(10√3 - 10t)²+(10)²

(10)² = (10√3-10t)² + (10)²

From here we get

t = √3sec

#answerwithquality

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