Question

a gun is held at an angle of 45 degree above the horizontal range of the projectile is 25 what is the muzzle velocity of the bullet fired from the given then how much it will take to reach the maximum height​

Answer 1

Answer:

solved

Explanation:

horizontal range  = u²sin2∝/g =  u²sin90°/g =  u²/g = u²/10 = 25

u = 5√10 m/s

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Horizontal range (R) = 25 meters.
• Angle of projection ( ${ \theta}$ ) = 45°

$\huge{\bold{\underline{Explanation:-}}}$

From the Range Formula:-

$\large{ \tt R = \dfrac{u^2 \sin 2 \theta}{g}}$

Substituting the values,

$\large{25 = \dfrac{u^2 \sin 2 \times 45}{10}}$

It becomes,

$\large{25 = \dfrac{u^2 \times \sin 90}{10}}$

as Sin 90° = 1,

then,

$\large{25 \times 10 = u^2}$

$\large{u = \sqrt{25 \times 10 }}$

$\large{\boxed{\boxed{ \tt u = 5 \sqrt{10} \: m/s}}}$

Additional information:-

As we know,

The total time of flight is (T = 2usinθ/g)

But question states that the Time taken by the projectile to reach the maximum height.

Therefore,

$\large{ \tt t_a = \dfrac{u^2 \sin \theta}{g}}$

Substituting the values,

$\large{t_a = \dfrac{(5 \sqrt{10})^2 \sin 45}{10}}$

Now,

$\large{t_a = \dfrac{25 \times 10 \times \dfrac{1}{ \sqrt{2}}}{10}}$

$\large{t_a = \dfrac{ \cancel{10} \times 25 \times \dfrac{1}{ \sqrt{2}}}{ \cancel{10}}}$

Now,

$\large{t_a = \dfrac{25}{ \sqrt{2}}}$

So,

$\large{ \sqrt{2} = 1.414}$

Substituting this on the above equation.

$\huge{\boxed{\boxed{ \tt t_a = 17.6 \: Seconds}}}$

Some useful relations:-

$\large{\bold{ \star \: R \tan \theta = 4H}}$

$\large{\bold{ \star \: H = \dfrac{gt^2}{8}}}$

Note:-

Here

• R = Maximum Range.
• H = Maximum Height.
• t = Time of flight.
• ta = Time of ascent.