When he lands his horizontal velocity is
<span>28 cos40 = 21.45 m/s </span>
<span>the time in flight comes from </span>
<span>x = Vht </span>
<span>58.8 = 21.45t </span>
<span>t = 2.74 seconds </span>
<span>his vertical velocity at landing is </span>
<span>28 sin40 = -18 m/s </span>
<span>his vertical velocity equation is </span>
<span>v = V0 - gt </span>
<span>-18 = V0 - 9.81(2.74) </span>
<span>V0 = -18 + 9.81(2.74) </span>
<span>V0 = 8.88 </span>
<span>his velocity magnitude was </span>
<span>v = (8.88^2 + 21.45^2)^½ </span>
<span>v = 23.2 m/s ANSWER </span>
<span>his initial direction was </span>
<span>tanθ = 8.88/21.45 </span>
<span>θ = 22.5 degrees above the horizontal ANSWER </span>
<span>to find the time to the flight apex from launch </span>
<span>v = gt </span>
<span>8.88 = 9.81t </span>
<span>t = 0.905 s </span>
<span>in 0.905 s Eddie has risen how far above the edge </span>
<span>y = ½(9.81)(0.905^2) </span>
<span>y = 4 m </span>
<span>the remainder of the flight is all drop and takes 2.74 - 0.905 = 1.85 seconds </span>
<span>in 1.85 seconds he drops </span>
<span>y = ½(9.81)(1.85^2) </span>
<span>y = 16.7 m </span>
<span>so the height from the edge to the landing point is </span>
<span>16.7 - 4 = 12.7 m ANSWER</span>
Answer:
In fluid dynamics, an object is moving at its terminal velocity if its speed is constant due to the restraining force exerted by the fluid through which it is moving. ... At this point the object ceases to accelerate and continues falling at a constant speed called the terminal velocity (also called settling velocity).
Answer:
Less than Mercury's
Explanation:
According to third Kepler's law, the square of the planet's orbital period is proportional to the cube of the average orbital radius of the planet's orbit. The constant of proportionality depends only on the mass of the star, recall that 51 Peg has the same mass as the Sun. Since the orbital period of this planet is less than Mercury's, its average orbital radius is less than Mercury's.
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