Question

A ball is thrown with initial velocity 10m/s making 60 degree angle with horizontal.find its range.

Answer 1

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• Let the вαℓℓ make an angle θ when it attains a velocity √50 after time t1.

• 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.

$hope \: its \: help \: u$

Answer 2

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• Let the вαℓℓ make an angle θ when it attains a velocity √50 after time t1.

• 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.

$hope \: its \: help \: u$