Motion of a projectile：練習問題

A projectile is launched from the origin with an initial speed of \(20 \text{ m/s}\) at an angle of \(45^\circ\) above the horizontal. Taking \(g = 10 \text{ m/s}^2\), which of the following is the Cartesian equation of its trajectory?

A particle is projected from a point at height \(h\) above horizontal ground with a horizontal velocity \(V\). Find the time taken for the particle to reach the ground.

A projectile is launched from a point on horizontal ground with speed \(u\) at an angle \(\alpha\) to the horizontal. If the horizontal range of the projectile is twice the maximum height reached, find the value of \(\tan \alpha\).

A projectile is fired from the origin with initial speed \(u\). The target is at point \((R, 0)\) on the horizontal plane. It is found that there are two possible angles of projection, \(\alpha_1\) and \(\alpha_2\), that allow the projectile to hit the target. If \(\alpha_1 = 15^\circ\), what is the value of \(\alpha_2\)?

A projectile is launched from ground level with speed \(u\) at an angle \(\alpha\). Let \(T\) be the total time of flight and \(H\) be the maximum height reached. Which of the following relationships is correct?

A projectile is launched with an initial speed \(u\) at an angle \(\alpha\) such that \(\tan \alpha = \frac{4}{3}\). Express the horizontal component of the velocity in terms of \(u\).

A particle is projected with speed \(u\) at an angle \(\alpha\). Show that the horizontal range \(R\) is maximized when \(\alpha = 45^\circ\) and state the maximum range.

A projectile is fired with initial speed \(V\) at an angle \(α\). Show that the time interval between the projectile being at height \(h\) on its upward path and being at height \(h\) on its downward path is \(\frac{2}{g}\sqrt{V^2\sin^2 α - 2gh}\).

A particle is projected from a point on horizontal ground with an initial speed of \(25 \text{ m s}^{-1}\) at an angle of \(35^\circ\) above the horizontal.

(a) Calculate the time of flight of the particle.
(b) Determine the horizontal range of the particle.
[Take \(g = 9.8 \text{ m s}^{-2}\)]

A projectile is fired from the origin with initial speed \(V\) at an angle \(\alpha\) to the horizontal. The equation of its trajectory is given by \(y = x \tan \alpha - \frac{gx^2}{2V^2 \cos^2 \alpha}\).

A target is located at a horizontal distance of \(40 \text{ m}\) and a height of \(10 \text{ m}\). If the initial speed is \(V = 25 \text{ m s}^{-1}\), find the two possible values of the angle of projection \(\alpha\). [Take \(g = 10 \text{ m s}^{-2}\)]