Question

At takeoff, the horizontal and vertical velocities of a high jumper are 2.0 and 3.9 m/s, respectively. What are the resultant velocity and angle of takeoff?

Answer 1

Answer on Question 63145, Physics, Mechanics

Question:

At takeoff, the horizontal and vertical velocities of a high jumper are 2.0 and $3.9 \, m/s$ , respectively. What are the resultant velocity and angle of takeoff?

Solution:

Here's the sketch of our task:

Here, $v$ is the resultant velocity of the high jumper; $v_{x}, v_{y}$ are the projections of the resultant velocity of the high jumper on axis $x$ and $y$ , respectively; $\theta$ is the angle of takeoff.

a) We can find the resultant velocity from the Pythagorean theorem:

$$v = \sqrt {v _ {x} ^ {2} + v _ {y} ^ {2}} = \sqrt {\left(2 . 0 \frac {m}{s}\right) ^ {2} + \left(3 . 9 \frac {m}{s}\right) ^ {2}} = 4. 3 8 \frac {m}{s}.$$

b) We can find the angle of takeoff from the triangle:

$$\tan \theta = \frac {v _ {y}}{v _ {x}},$$ $$\theta = \tan^ {- 1} \left(\frac {v _ {y}}{v _ {x}}\right) = \tan^ {- 1} \left(\frac {3 . 9 m / s}{2 . 0 m / s}\right) = 6 2. 8 {}^ {\circ}.$$

Answer:

a) $v = 4.38 \, m/s$

b) $\theta = 62.8{}^{\circ}$

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