Question

Trying to escape his pursuers, a secret agent skis off a slope inclined at 30° below the horizontal at 55 km/h. to survive and land on the snow 100 m below, he must clear a gorge 43 m wide. does he make it? ignore air resistance.

Answer 1

I shall draw this out as it’s a two dimensional kinematics problem. Though, if I do have an error with my work, then it’s just with the units of conversion, yet I believe my calculations are accurate.

Answer 2

The agent made it since the range of his motion is greater than the width of the gorge.

The given parameters;

• inclination of the slope, Ф = 30°

• velocity of the agent, v = 55 km/h

• vertical distance below the snow, h = 100 m

• width of the gorge the agent must cross, w = 43 m

Assume downward motion to be negative.

The initial vertical component of the velocity in m/s;

$v_0_y = v_0 \times sin(30)\\\\v_o_y = - 55 \ km/h \times sin(30) = -27.5 \ km/h = -7.64 \ m/s$

Determine the final vertical velocity of the agent;

$v_y_f^2 = v_0_y^2 - 2gh\\\\v_y_f^2 = (-7.64)^2 - 2(9.8)(-100)\\\\v_y_f^2 = 2018.37\\\\v_y_f = \sqrt{2018.37} \\\\v_y_f = -44.93 \ m/s$

Determine the time of motion;

$v_y_f = v_0y - gt\\\\-44.93 = -7.64 - 9.8t\\\\9.8t = 44.93 - 7.64\\\\9.8t = 37.29\\\\t = \frac{37.29}{9.8} = 3.81 \ s$

Determine the horizontal component of the initial velocity;

$v_0_x = v_0 \times cos(\theta)\\\\v_0_x = 55 \ km/h \times cos(30) = 47.63 \ km/h = 13.2 \ m/s$

Determine the range of the of the agent's motion;

$X = v_0_x \times t\\\\X = 13.2\ m/s \times 3.81 \ s\\\\X = 50.3 \ m$

Thus, the agent made it since the range of his motion is greater than the width of the gorge.

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