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7nadin3 [17]
3 years ago
10

A long-jumper jumps at a speed of 9.65 m/s at an angle of 21.00 to the ground, 2.00 m before the edge of a ditch which is 1.50 m

deep and infinitely long. Assume projectile motion for the path of her jump. (a) Find the time of her jump. (b) Find the distance from the edge of the cliff to the point of her impact with the ground. (c) Find the maximum height of her jump above the bottom of the ditch. (d) Find her velocity in vector notation, just before the point of her impact with the ground.
Physics
1 answer:
Tcecarenko [31]3 years ago
7 0

Answer:

a) t=0.9935s

b) d = 6.95m

c) Hmax=2.1m

d) Vf = (9.009,-6.48) m/s

Explanation:

First of all, the we calculate the components of the initial velocity:

V_{ox}=Vo*cos(21)=9.009m/s

V_{oy}=Vo*sin(21)=3.46m/s

For the time of the jump:

Y_{f}=Y_{o}+V_{oy}*t-\frac{g*t^{2}}{2}  Solving the quadratic equation for t, we get:

t = 0.9935s

For the distance of her impact, we will need the time we just calculated:

D=X_{o}+V_{ox}*t=6.95m

To know the maximum height:

V_{fy}=V_{oy}-g*t_{Hmax}

t_{Hmax}=\frac{V_{oy}}{g} =0.346s  Using this value, we calculate the maximum height:

Hmax=Y_{o}+V_{oy}*t_{Hmax}-\frac{g*t_{Hmax}^{2}}{2}=2.1m

Finally, for the final velocity, we use the time of the jump t=0.9935s:

V_{fx}=V_{ox}=9.009m/s

V_{fy}=V_{oy}-g*t=-6.48m/s

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A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.6 m/s . Two seconds later t
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Answer:

A) t = 7.0 s    

B) x = 25 m  

C) v = 10 m/s

Explanation:

The equations for the position and velocity of an object traveling in a straight line is given by the following expressions:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A)When both friends meet, their position is the same:

x bicyclist = x friend

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

If we place the center of the frame of reference at the point when the bicyclist starts following his friend, the initial position of the bicyclist will be 0, and the initial position of the friend will be his position after 2 s:

Position of the friend after 2 s:

x = v · t

x = 3.6 m/s · 2 s = 7.2 m

Then:

1/2 · a · t² = x0 + v · t       v0 of the bicyclist is 0 because he starts from rest.

1/2 · 2.0 m/s² · t² = 7.2 m + 3.6 m/s · t

1  m/s² · t² - 3.6 m/s · t - 7.2 m = 0

Solving the quadratic equation:

t = 5.0 s

It takes the bicyclist (5.0 s + 2.0 s) 7.0 s to catch his friend after he passes him.

B) Using the equation for the position, we can calculate the traveled distance. We can use the equation for the position of the friend, who traveled over 7.0 s.

x = v · t

x = 3.6 m/s · 7.0 s = 25 m

(we would have obtained the same result if we would have used the equation for the position of the bicyclist)

C) Using the equation of velocity:

v = a · t

v = 2.0 m/s² · 5.0 s = 10 m/s

8 0
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