A projectile is shot from the edge of a cliff 245 m above ground level with an i

Question

A projectile is shot from the edge of a cliff 245 m above ground level with an initial speed of vo=125 m/s at an angle of 35.0° with the horizontal, as shown in the figure.

(a) Determine the time taken by the projectile to hit point P at ground level.

(b) Determine the distance X of of point P from the base of the vertical cliff.

(c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Assume the positive directions are upward and to the right.)

(d) At the instant just before the projectile hits point P, what is the magnitude of the velocity?

(e) At the instant just before the projectile hits point P, what is the angle made by the velocity vector with the horizontal?
° below the horizontal

(f) Find the maximum height above the cliff top reached by the projectile.

Explanation / Answer

To answer this question, write the motion equation in eachdimension, horizontal and vertical    final position = initialposition + init velocity * time + (1/2) at2 Horizontal:            X    =    0   +   125 cos35 * t    +      0 Vertical:           0   =    245   +    125 sin35 * t    -    (1/2) * 9.8 * t2 so, 4.9 t^2 - 71.69 t - 245 = 0 The vertical equation is a quadratic, solve it and youget        t  =      17.48 seconds     is how long it takes to hitthe ground ............ans of [a]    and     X = 125cos35 * 17.48 =    1789.84 meters   = 1.789 km.................ans of [b] Components of velocity? Just use    final vel = initvel + at      horizontal:     final vel = 125cos35 + 0   = 102.39 m/s      vertical:       final vel  =   125 sin35   - 9.8 *17.48 =    -99.60 m/s .........ans 0f [c] magnitude...   (102.392 + 99.62 )1/2   = 142.84 m/s ...................................ans of [d] direction    tan = 99.6/102.39 =    44.2 deg below horizontal ..........ans of [e] maximum height = 1789.84 meters- 245 m = 1544.84 m above the cliff top reached by the projectile. .....ans of [f] Components of velocity? Just use    final vel = initvel + at      horizontal:     final vel = 125cos35 + 0   = 102.39 m/s      vertical:       final vel  =   125 sin35   - 9.8 *17.48 =    -99.60 m/s .........ans 0f [c] magnitude...   (102.392 + 99.62 )1/2   = 142.84 m/s ...................................ans of [d] direction    tan = 99.6/102.39 =    44.2 deg below horizontal ..........ans of [e] maximum height = 1789.84 meters- 245 m = 1544.84 m above the cliff top reached by the projectile. .....ans of [f]

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