PLEASE ADD DETAILS, I REALLY WANT TO LEARN THIS MATERIAL NOT JUST ANSWERS. THANK

Question

PLEASE ADD DETAILS, I REALLY WANT TO LEARN THIS MATERIAL NOT JUST ANSWERS. THANK YOU.

A high diver leaves the end of a 5.0 m high diving board and strikes the water 1.3 s later, 2.8 m beyond the end of the board. Considering the diver as a particle, determine the following A projectile is shot from the edge of a cliff h = 245 m above ground level with an initial speed of V0 = 115 m/s at an angle of 37.0 degree with the horizontal, as shown in the figure below. Determine the time taken by the projectile to hit point P at ground level. Determine the range X of the projectile as measured from the base of the cliff. At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Take up and to the right as positive directions.) What is the the magnitude of the velocity? m/s What is the angle made by the velocity vector with the horizontal? Find the maximum height above the cliff top reached by the projectile

Explanation / Answer

Number 1)

Part A)

In the vertical direction, start with d = vot + .5at2

-5 = vo(1.3) + (.5)(-9.8)(1.3)2

vo = 2.52 m/s

In the x direction, start with d = vt

2.8 = v(1.3)

v = 2.15 m/s

The initial velocity is found by combining the components using the Pythagorean Theorem...

v2 = (2.15)2 + (2.52)2

vo = 3.32 m/s

The angle is from the tangent function

tan(angle) = 2.52/2.15

angle = 49.5o above the horizontal

Part B)

The max height is from the vertical information where the velocity at the max height is zero in that direction.

Apply vf2 = vo2 + 2ad

0 = (2.52)2 + 2(-9.8)(d)

d = .324 m

Total height = 5 + .324 = 5.324 m

Part C)

We need the final vertical velocity when hitting the water

Apply vf2 = vo2 + 2ad

vf2 = 2.522 + 2(9.8)(5)

vf = 10.2 m/s

Now combine that with the Pythagorean Theorem for the components...

vf2 = (10.2)2 + (2.15)2

vf = 10.42 m/s

The angle for the tangent function is

tan(angle) = 10.2/2.15

angle = 78.1o below the horizontal

Number 2)

Part A)

In the vertical direction, the initial velocity = v(sin angle)

That is 115(sin 37) = 69.21 m/s

Then find the final y velocoty

vf2 = vo2 + 2ad

vf2 = (69.21)2 + 2(9.8)(245)

vf = -97.9 m/s (Negative assigned since it will be downward)

Then you can find the time to do that travel

vf = vo + at

-97.9 = 69.21 + (-9.8)(t)

t = 17.1 sec

Part B)

In the x direction, the initial velocity = v(cos angle)

115(cos 37) = 91.8 m/s

Then apply d = vt in the horizontal

d = (91.8)(17.1)

d = 1570 m which is 1.57 km

Part C)

Already found in earlier parts

vy final = -97.9 m/s

vx final = 91.8 m/s

Part D)

Net velocity by the Pythagorean Theorem

v2 = (91.8)2 + (97.9)2

v = 134.2 m/s

Part E)

The angle is from the tangent function

tan(angle) = 97.9/91.8

angle = 46.8o Below the horizontal

Part F)

In the vertical direction, apply vf2 = vo2 + 2ad

0 = (69.21)2 + 2(-9.8)(d)

d = 244.4 m

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