Question

A projectile thrown with a velocity of 10m/s at an angle 60° with horizontal. The interval between the moment when speed is underoot 5g m/ s

Answer 1

Answer:

Explanation:

It is given that at time t=0, the velocity of the projectile is 10m/s and it makes an angle of 60 degree with the horizontal.

The horizontal component of the velocity of the projectile remains constant.

Therefore,

√50 cosθ= 10 cos 60

√50 cosθ= 5

cosθ= 5/√50

cosθ= 1/√2

Therefore θ= 45 degree.

The vertical component of the velocity √50 is

=√50 sin45

=√50/√2

=5 m/sec

The vertical component of the velocity at the start was= 10 sin60

= 10x (√3/2) = 5 √3

The time t1 taken for the vertical component of the velocity to reduce from 5 √3 m/s to 5m/s is=

5= 5 √3 - g t1 ( where g is the acceleration due to the gravity= 10m/s^2)

t1= 5(√3 -1)/10

t1= (√3 -1)/2

The same velocity √50 will be attained again at time t2, when the vertical component of the velocity is again 5m/s , however, in the downward direction with the horizontal component of the velocity being the same throughout.

Or,

-5 = 5 √3 - g t2

t2=(√3 +1)/2

Therefore t2-t1 =

1 sec.

Answer 2

Given: A projectile thrown at an angle of 60° with horizontal with a velocity of 10m/s

To find: the interval between the moment when speed is √5g m/ s.

Solution: The speed at which the projectile was thrown was 10m/s

Its horizontal component will be 10cos60°= 5m/s which will remain constant throughout the motion.

When the speed is √5g we can write it as

(√5g)^2= v^2+ 5^2

v^2= 25

v= 5m/s

Now, let the projectile takes t time to travel from this point to the top at which the speed will be 0. By symmetry, we can say that it will take equal time to return to the same height.

Therefore, the total time taken by particle to cover the distance between this point will be 2t

now we know that v= u+at

0= 5-gt

t= 0.5s

The total time will be 2t= 1s.