A projectile is launched with velocity \(100 \mathrm{~ms}^{-1}\) at \(60^{\circ}\) to the horizontal. Atmospheric drag is negligible. Find the maximum height attained and the range. What other angle of launch would give the same range? Find the time of flight in each of the two cases.

Projectile motion can be understood as two separate motions—horizontal motion with constant velocity and vertical motion with constant acceleration—that occur simultaneously. Horizontal motion is described by the equation:$$x = v_{0x}t$$ Vertical motion is described by the equations:$$y = v_{0y}t - \frac{1}{2}gt^2$$$$v_{y} = v_{0y} - gt$$In this problem, atmospheric drag is negligible, so motion is considered ideal projectile motion. The time of flight is the same for both horizontal and vertical motions.

Given the initial velocity \(v_0 = 100 \mathrm{~ms}^{-1}\) and launch angle \(\theta = 60^{\circ}\), we can find the horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) components of the initial velocity using trigonometry:$$v_{0x} = v_0\cos{\theta}$$$$v_{0y} = v_0\sin{\theta}$$Using these formulas, we calculate the initial velocity components:$$v_{0x} = 100 \cos(60^{\circ}) = 50 \mathrm{~ms}^{-1}$$$$v_{0y} = 100 \sin(60^{\circ}) = 86.60 \mathrm{~ms}^{-1}$$

To find the maximum height, we can use the vertical motion equation for final velocity (\(v_{y}\)) and solve for \(y\): $$v_{y} = v_{0y} - gt$$At the maximum height, the vertical velocity is zero (\(v_y = 0\)). Substituting the values and solving for \(y\):$$0 = 86.60 \mathrm{~ms}^{-1} - (9.8 \mathrm{~ms^{-2}})(t)$$$$t = 8.84 \mathrm{s}$$Now, using the vertical displacement equation, we find the maximum height:$$y = v_{0y}t - \frac{1}{2}gt^2$$$$y = (86.60 \mathrm{~ms^{-1}})(8.84 \mathrm{s}) - \frac{1}{2}(9.8 \mathrm{~ms^{-2}})(8.84 \mathrm{s})^2$$$$y = 382.23 \mathrm{m}$$The maximum height attained is 382.23 m.

Next, we'll use the horizontal motion equation to find the range of the projectile using the time of flight from the vertical motion:$$x = v_{0x}t$$$$x = (50 \mathrm{~ms^{-1}})(8.84 \mathrm{s})$$$$x = 442.00 \mathrm{m}$$The range of the projectile is 442.00 m.

Another angle of launch will produce the same range if the horizontal component of the initial velocity is equal. Since the initial velocity is constant, the alternative angle must be complementary to the given angle. So, the alternative angle is $$90^{\circ} - 60^{\circ} = 30^{\circ}$$. The time of flight for this alternative angle can be found using the vertical motion equation:$$t = \frac{2v_{0y'}}{g}$$Where \(v_{0y'}\) is the vertical component of the initial velocity for the alternative angle. Calculate the vertical component and the time of flight:$$v_{0y'} = v_0\sin{30^{\circ}} = 50 \mathrm{~ms}^{-1}$$$$t' = \frac{2(50 \mathrm{~ms}^{-1})}{9.8 \mathrm{~ms^{-2}}}$$ $$t' = 10.20 \mathrm{s}$$The alternative angle of launch is \(30^{\circ}\), and the time of flight for this angle is 10.20 s. To summarize: - Maximum height: 382.23 m - Range: 442.00 m - Alternative angle of launch: \(30^{\circ}\) - Time of flight for original angle: 8.84 s - Time of flight for alternative angle: 10.20 s