Exemplo de Physics, 5.3 Projectile Motion

Calculating Projectile Motion: Hot Rock Projectile

Problema e solução em inglês, como no livro original.

Suppose a large rock is ejected from a volcano, as illustrated in Figure 5.31, with a speed of $25.0 m / s$ and at an angle $35 °$ above the horizontal. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path.

A volcano is shown with labels illustrating the projectile motion of rock shooting out of the volcano. Vector v o equals twenty-five meters per second and points from the volcano opening upward. Vector v curves downward to the ground and forms an angle of thirty-five degrees with a horizontal plane at the height of the volcano opening. The distance between the horizontal plane and the ground is labeled as negative twenty meters.

Figure 5.31 The diagram shows the projectile motion of a large rock from a volcano.

Resolva no papel primeiro. Depois abra a solução um passo de cada vez e pare assim que conseguir terminar sozinho.

Estratégia
Breaking this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the time. The time a projectile is in the air depends only on its vertical motion.
Solução
While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

$y = y_{0} + v_{0 y} t - \frac{1}{2} g t^{2} .$
If we take the initial position $y_{0}$ to be zero, then the final position is $y = - 20.0 m .$ Now the initial vertical velocity is the vertical component of the initial velocity, found from

$v_{0}_{y} = v_{0} sin θ_{0} = (25.0 m/s) (sin 35^{\circ}) = 14.3 m/s .$
5.9

Substituting known values yields

$- 20.0 m = (14.3 m/s) t - (4.90 {m/s}^{2}) t^{2} .$

Rearranging terms gives a quadratic equation in $t$

$(4.90 {m/s}^{2}) t^{2} - (14 .3 m/s) t - (20.0 m) = 0 .$

This expression is a quadratic equation of the form $a t^{2} + b t + c = 0$ , where the constants are a = 4.90, b = –14.3, and c = –20.0. Its solutions are given by the quadratic formula

$t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} .$

This equation yields two solutions t = 3.96 and t = –1.03. You may verify these solutions as an exercise. The time is t = 3.96 s or –1.03 s. The negative value of time implies an event before the start of motion, so we discard it. Therefore,

$t = 3.96 s.$

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