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

Explanation:

Part (a)

Step1: Find vertical component of velocity

The initial vertical velocity \( v_{0y} = v_0 \sin\theta \), where \( v_0 = 50 \, \text{m/s} \) and \( \theta = 60^\circ \). 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 time) and \( t = \frac{2v_{0y}}{g} \). Substituting \( v_{0y} \approx 43.30 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \): \( t = \frac{2 \times 43.30}{9.8} \approx 8.84 \, \text{s} \).

Step1: Find horizontal component of velocity

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

Step2: Calculate horizontal distance (range)

Range \( R = v_{0x} \times t \), where \( t \) is time of flight from part (a) (\( \approx 8.84 \, \text{s} \)). So \( R = 25 \times 8.84 = 221 \, \text{m} \) (approx).

Step1: At maximum height, vertical velocity is 0

Use the equation \( v_y^2 = v_{0y}^2 - 2g(y - y_0) \). At max height, \( v_y = 0 \), so \( 0 = v_{0y}^2 - 2g(h - 0) \) (let \( h \) be max height, \( y_0 = 0 \)).

Step2: Solve for \( h \)

Rearranging: \( h = \frac{v_{0y}^2}{2g} \). Substituting \( v_{0y} \approx 43.30 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \): \( h = \frac{(43.30)^2}{2 \times 9.8} \approx \frac{1874.89}{19.6} \approx 95.66 \, \text{m} \).

Answer:

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

Part (b)