Question

a projectile is launched at an angle of 45 degree with a velocity of 250m/s. If air resistance is neglected. the magnitude of the horizontal velocity of the projectile at the time it reaches maximum altitude is equal to?​

Answer 1

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a projectile is launched at an angle of 45 degree with a velocity of 250m/s. If air resistance is neglected. the magnitude of the horizontal velocity of the projectile at the time it reaches maximum altitude is equal to?

☆.an angle of 45 degree of velocity of = 250m/s.

☆.they are asking what is your altitudes is equal

to.. = ?

DEFINITION:-

✒To describe projectile motion completely, we must include velocity and acceleration, as well as displacement. We must find their components along the x- and y-axes.

✒ Let’s assume all forces except gravity (such as air resistance and friction, for example) are negligible. Defining the positive direction to be upward, the components of acceleration are then very simple:

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SOLUTION:-

The range (R) of the projectile is the horizontal distance it travels during the motion.

Now, s = ut + ½ at2

Using this equation vertically, we have that a = -g (the acceleration due to gravity) and the initial velocity in the vertical direction is usina (by resolving).

Hence:

Hence:y = utsina - ½ gt2       (1)

Using the equation horizontally:

x = utcosa           (2)

Remember, there is no acceleration horizontally so a = 0 here.

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ANSWER:-

Applying this equation vertically, when the particle hits the ground:

0 = 25Tsin30 - ½ gT2 (Where T is the time of flight) 

Therefore, T(25sin30 - ½ gT) = 0

So T = 0 or T = (50sin30)/g

Therefore the time of flight is 2.55s (3sf) 

b) The range can be found working out the horizontal distance travelled by the particle after time T found in part.

✒

s=ut−21gt2

s=ut−21gt2 gt=2sinθ

s=ut−21gt2 gt=2sinθ t=g2sinθ

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(a) .s = ut + ½ at2

Applying this equation horizontally:

R = 25Tcos30

= 25 × 2.55 × 0.866

= 55.231...  

✒Therefore the horizontal distance travelled is 55.2m(3sf).

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☆. Horizontal Motion:-

  

✒R=ucosθ×g2usinθ

 R=gu2sin2θ

☆Vertical Motion:-

✒v2=u2−2as

 0=u2sin2θ−2gH

 H=2gu2sin2θ.

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HENCE ,VELOCITY IS

At highest point :

✒At highest point :velocity = ucosθ =?

✒At highest point :velocity = ucosθ = 250 × cos45°

✒At highest point :velocity = ucosθ = 250 × cos45°= 250 × 1/√2

✒At highest point :velocity = ucosθ = 250 × cos45°= 250 × 1/√2= 177.3m/s

SO, HENCE PROVED 177.3M/S .

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