R = u2sin2A = (20)2Sin2(45) = 40Sin90 how this last step comes out to be? g9.89.8
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Range of a two dimensional project in Earth's gravitational field.
Horizontal speed = u Cos A
vertical speed initial = u Sin A
vertical velocity = v= u Sin A - g t v = 0, when t = u sin A/g
The projectile lands when t = 2 * u sin A / g.
Range = horizontal distance travelled = u CosA * 2u Sin A / g
R = u^2 Sin 2A / g
If u =20 m/s, and A = 45 deg for maximum range, for a given speed u:
R = 20 * 20 Sin(2*45) / 10 as g = 10 m/sec/sec
R = 20*2 Sin90 = 40 meters
Horizontal speed = u Cos A
vertical speed initial = u Sin A
vertical velocity = v= u Sin A - g t v = 0, when t = u sin A/g
The projectile lands when t = 2 * u sin A / g.
Range = horizontal distance travelled = u CosA * 2u Sin A / g
R = u^2 Sin 2A / g
If u =20 m/s, and A = 45 deg for maximum range, for a given speed u:
R = 20 * 20 Sin(2*45) / 10 as g = 10 m/sec/sec
R = 20*2 Sin90 = 40 meters
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