Question

What initial launch speed is required to allow a projectile to have a range of 300 m if it is launched at an angle of 30◦ above the horizontal

Answer 1

Answer: 58.85m/s

Explanation:

Answer 2

Answer:

58.85m/s

Explanation:

Given:

Range=300m

Angle=$30^{o}$

To find:

Initial speed

Solution:

The only motion an item in the air may experience is projectile motion, which is caused by gravity. A projectile that is fired into the air close to the surface of the Earth and, assuming that air resistance is minimal, travels along a curved path, or more precisely a parabolic path, as a result of the force of gravity The science of such motion is known as ballistics. Gravity is the only force acting on a projectile; as a result, the bullet accelerates downward. An external force is not required to keep the object moving horizontally because of its initial inertia. When the projectile motion is on a bigger scale, it also necessitates taking into account other external forces such friction from aerodynamic drag or internal propulsion.

x(t) --> $x_o+v_ot+\frac{1}{2} at^{2}$

y(t) --> $y_o+v_ot+\frac{1}{2} at^{2}$

vy(t)-->$v_osin\theta+at$

$\theta - > \frac{v_o}{2g}-gt$

$t- > \frac{v_o}{2g}$

Thus the total time for projectile to fall - 2t

$2t- > \frac{2v_0}{2g}- > \frac{v_o}{g}$

$x(t)- > x_o+v_ot+\frac{1}{2} at^{2} \\300=v_o(\frac{v_o}{g} )cos\theta\\300=\frac{v_o^{2} }{y} cos\theta\\300=(v_o)^{2}\frac{\sqrt{3} }{2} \\v_o=\sqrt{\frac{6000}{\sqrt{3} } }\\= 58.85m/s$

Projectile motion and time of flight and maximum height

https://brainly.in/question/201038

For projectile motion Vi=2m/s and angle 30,Vy​

https://brainly.in/question/31633425

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