Question

The drawing shows a skateboarder moving at 4.8 m/s along a horizontal section of a track that is slanted upward by 48° above the horizontal at its end, which is 0.54 m above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height H to which she rises above the end of the track.

Answer 1

Question #16215

Let $\nu_{0}$ denote the velocity at horizontal part of the track, and $\nu_{0}^{\prime}$ denote the velocity at the end of the upward part of the track. First, let's find $\nu_{0}^{\prime}$ , using the law of conservation of energy:

$$\frac {m v _ {0} ^ {2}}{2} = \frac {m v _ {0} ^ {\prime 2}}{2} + m g h \Rightarrow v _ {0} ^ {\prime 2} = v _ {0} ^ {2} - 2 g h$$

Then, for the movement from the end of the upward part of the track:

$$S _ {x} = v _ {0} ^ {\prime} \cos \varphi t$$ $$S _ {y} = v _ {0} ^ {\prime} \sin \varphi t - \frac {g t ^ {2}}{2}$$

Find the maximum of $S_{y}$ : $\frac{dS_{y}}{dt} = v_{0}'\sin \varphi t - gt = 0 \Rightarrow T = \frac{v_{0}'\sin \varphi}{g}$ (3)

At this moment of time, the skateboarder will reach the maximum height.

Plugging (3) into (2), obtain: $H_{\text{max}} = \frac{v_0'^2 \sin^2 \varphi}{2g}$ . Plugging (1) into latter formula, finally obtain:

$$H _ {m a x} = \frac {\left(v _ {0} ^ {2} - 2 g h\right) \sin^ {2} \varphi}{2 g} = 0. 3 5 m.$$