An arrow with a mass of 30g is shot straight up to a height of 130m what is the arrow's initial velocity Please help?

Answer 1

Initial speed when it left the bow = 'V'

Final speed when it reached peak height = zero

Average speed during the climb = 1/2 V

Distance of the climb = (average speed in the climb) x (time in the climb)

130 = (1/2 V) x (time in the climb)

Time = 130 / (1/2 V) = (260) / V

But we also know that the upward speed decreases by 9.8 m/s every second,

because of the acceleration of gravity. So the time spent in the climb is (V / 9.8).

V / 9.8 = 260 / V

Multiply both sides by V:

V2 / 9.8 = 260

Multiply both sides by 9.8:

V2 = 9.8 x 260 = 2,548

V = sqrt(2548) = 50.48 meters per second (rounded)

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One important thing to mention: The mass of the arrow doesn't matter. An arrow, a bullet, a rock,

a cannonball, or a rocket ship, all do exactly the same thing if the initial velocity is the same.

That's the way gravity works. If there's any difference, it's only on account of air resistance.

Answer 2

To find the initial velocity of the arrow, you can use the kinematic equation: ( v^2 = u^2 + 2as ), where:

• ( v = 0 ) (final velocity at the top)
• ( u ) = initial velocity
• ( a = -9.81 , m/s^2 ) (acceleration due to gravity)
• ( s = 130m ) (height)

Rearranging the equation to solve for ( u ), you get ( u = \sqrt{v^2 - 2as} ). Plugging in the values, ( u = \sqrt{0 - 2*(-9.81)*130} = 50.86 , m/s ). Therefore, the arrow's initial velocity is approximately 50.86 m/s.