Question

1. a particle is launched from the ground with a speed of 50 m/s at an angle and 60° above the horizontal. the particle lands back on the ground after some time. you can use the magnitude of the acceleration due to gravity is g = 9.8 m/s² in the following problems.

a) (5pts) how long does it take the object to hit the ground?
b) (5pts) how far does the object travel?
c) (5pts) what is the maximum height above the ground reached by the object?
d) (5pts) at its maximum height, what is the velocity of the object and what is the acceleration of the object? these are vectors. please write them in vector form.

Explanation:

Part (a)

Step1: Identify vertical velocity component

The initial vertical velocity \( v_{0y} = v_0 \sin\theta \), where \( v_0 = 50 \, \text{m/s} \), \( \theta = 60^\circ \). So \( v_{0y} = 50 \sin 60^\circ = 50 \times \frac{\sqrt{3}}{2} \approx 43.30 \, \text{m/s} \).

Step2: Use vertical motion equation for time of flight

The vertical displacement \( y - y_0 = 0 \) (starts and lands on ground). The equation is \( y = y_0 + v_{0y}t - \frac{1}{2}gt^2 \). Substituting \( y = y_0 = 0 \), we get \( 0 = v_{0y}t - \frac{1}{2}gt^2 \). Factoring out \( t \): \( t(v_{0y} - \frac{1}{2}gt) = 0 \). Solutions are \( t = 0 \) (launch) and \( t = \frac{2v_{0y}}{g} \). Plugging in values: \( t = \frac{2 \times 43.30}{9.8} \approx 8.84 \, \text{s} \).

Step1: Identify horizontal velocity component

Horizontal velocity \( v_{0x} = v_0 \cos\theta = 50 \cos 60^\circ = 25 \, \text{m/s} \) (constant, no air resistance).

Step2: Use horizontal motion for range

Range \( R = v_{0x} \times t_{\text{flight}} \). From part (a), \( t_{\text{flight}} \approx 8.84 \, \text{s} \). So \( R = 25 \times 8.84 = 221 \, \text{m} \) (or using range formula \( R = \frac{v_0^2 \sin 2\theta}{g} \), \( \sin 120^\circ = \frac{\sqrt{3}}{2} \), \( R = \frac{50^2 \times \frac{\sqrt{3}}{2}}{9.8} \approx 221 \, \text{m} \)).

Step1: At maximum height, vertical velocity \( v_y = 0 \)

Use the equation \( v_y^2 = v_{0y}^2 - 2g\Delta y \). Solving for \( \Delta y \) (maximum height \( h \)): \( h = \frac{v_{0y}^2}{2g} \).

Step2: Substitute \( v_{0y} \approx 43.30 \, \text{m/s} \)

\( h = \frac{(43.30)^2}{2 \times 9.8} = \frac{1874.89}{19.6} \approx 95.66 \, \text{m} \).

Answer:

\(\approx 8.84 \, \text{seconds}\)

Part (b)