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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.

Answer :

lcmendozaf

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:

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

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

For the time of the jump:

[tex]Y_{f}=Y_{o}+V_{oy}*t-\frac{g*t^{2}}{2}[/tex]  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:

[tex]D=X_{o}+V_{ox}*t=6.95m[/tex]

To know the maximum height:

[tex]V_{fy}=V_{oy}-g*t_{Hmax}[/tex]

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

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

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

[tex]V_{fx}=V_{ox}=9.009m/s[/tex]

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

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