Question

Answer asap giving max points and brainliest A projectile is launched at an angle of 30° and travels a distance of 400 meters. Gravitational acceleration equals 9.8 m/sec2 calculate the initial velocity.
answer choices:
67
87
57

Answer 1

Answer:

$\sf u=67\:ms^{-1}$

Step-by-step explanation:

Assuming the distance traveled is the horizontal distance.

Horizontal Range Formula

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

where:

• R = horizontal range
• u = initial velocity
• $\theta$ = angle of initial velocity
• g = acceleration due to gravity

Given:

• R = 400 m
• $\theta$ = 30°
• g = 9.8 m/s²

Substituting the given values into the formula and solving for u:

$\implies \sf 400=\dfrac{u^2 \sin 60^{\circ}}{9.8}$

$\implies \sf 3920=u^2 \sin 60^{\circ}$

$\implies \sf 3920=\left(\dfrac{\sqrt{3}}{2}\right)u^2$

$\implies \sf u^2=\dfrac{7840}{\sqrt{3}}$

$\implies \sf u=\sqrt{\left(\dfrac{7840}{\sqrt{3}} \right) }$

$\implies \sf u=67.2787196...$

$\implies \sf u=67\:ms^{-1}\:(nearest\:whole\:number)$

Answer 2

Answer:

67 m/s

Step-by-step explanation:

Formula for range :

• Range = u²sin2θ / g

1. u = initial velocity
2. θ = angle of launch
3. g = gravitational acceleration

Solving with the given values :

• 400 = u² x sin60° / 9.8
• u² x √3/2 = 3920
• u² = 7840/√3
• u² = 4526.4
• u = 67 m/s (approximately)