Question

a projectile is shot at an inclination of 45 frin tge horizontial with a speed of 250 m/s. how far will it travek ub the horizontal direction

Answer 1

Answer:

6250 m  

Explanation:

When an object is projected into the air, the distance along the horizontal direction is called the Range.

The Range of a projectile is expressed as;

                                   $R = \frac{u^{2}sin2\alpha }{g}$

Where,

R is the range of the projectile

α is the angle of inclination with the horizontal

g is the acceleration due to gravity = 9.8 m/s ≈ 10 m/s

Given; α =45° , u = 250 m/s

                                $R = \frac{250^{2}sin2(45)}{10}$

                                $R = \frac{62500sin90}{10}$

                                $R = \frac{62500}{10}$

                                R =  6250 m

The range is 6250 m                              

Answer 2

In a projectile motion, the given object travel 6377.55 m in the horizontal direction.

 

In a projectile motion, the distance of the object along the horizontal direction is called the Range.  

The Range of a projectile                    

$\bold {R = \dfrac {u^2 sin2\alpha }{g}}$

Where,  

R - range of the projectile

u - initial speed = 250 m/s  

α - angle of inclination with the horizontal = 45°  

g - gravitational acceleration = 9.8 m/s  

Put the values in the formula,

$\bold {R = \dfrac {(250)^2 sin2(45) }{9.8}}\\\\\bold {R = \dfrac {62500\ sin 90 }{9.8}}\\\\\bold {R =6377.55}$  Since, sin 90 = 1

Therefore, in a projectile motion, the given object travel 6377.55 m in the horizontal direction.

To know more about projectile motion,  

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