Question

How best can you calculate the time a coin was in the air when it was projected at an angle of 53° above the horizontal from a point 15m above the ground level and it reaches the ground at a horizontal distance of 40m from the launch point?

Answer 1

To calculate the time a coin was in the air, we can use the equations of motion. In this case, since the coin is projected at an angle above the horizontal, we need to break down the motion into horizontal and vertical components.

Let's analyze the vertical motion first:

1. Initial vertical velocity (Vy0): Since the coin is projected at an angle of 53° above the horizontal, we can calculate the initial vertical velocity using the formula Vy0 = V * sin(θ), where V is the initial velocity and θ is the angle of projection.

   Given that we don't have the initial velocity, we need to find it. To do that, we can use the horizontal motion component.

Now let's analyze the horizontal motion:

2. Initial horizontal velocity (Vx0): We can calculate the initial horizontal velocity using the formula Vx0 = V * cos(θ).

   However, we still don't have the initial velocity (V), so we need to find it using the given information.

3. Time of flight (T): The time the coin is in the air can be determined by calculating the time it takes for the horizontal displacement to occur. We can use the formula: T = distance / Vx0, where distance is the horizontal distance traveled.

Now, let's find the missing values step by step:

1. Initial vertical velocity (Vy0): Vy0 = V * sin(θ) Given that θ = 53°, we need to find V.

2. Initial horizontal velocity (Vx0): Vx0 = V * cos(θ) Given that θ = 53°, we need to find V.

3. Time of flight (T): T = distance / Vx0 Given distance = 40m, we need to find Vx0.

To find V, we'll use the information that the coin is projected from a point 15m above the ground. Assuming the initial and final heights are the same (since the coin will be at the ground level), we can use the vertical motion equation:

4. Final vertical displacement (Δy): Δy = Vy0 * T + (1/2) * g * T^2 Here, g represents the acceleration due to gravity, approximately 9.8 m/s^2.

   Given that Δy = 15m, we can solve for T using the quadratic equation.

Let's calculate step by step:

1. Initial vertical velocity (Vy0): Vy0 = V * sin(θ) Vy0 = V * sin(53°)

2. Initial horizontal velocity (Vx0): Vx0 = V * cos(θ) Vx0 = V * cos(53°)

3. Time of flight (T): T = distance / Vx0 T = 40m / (V * cos(53°))

4. Final vertical displacement (Δy): Δy = Vy0 * T + (1/2) * g * T^2 15m = V * sin(53°) * T + (1/2) * 9.8 m/s^2 * T^2

Now, we have two equations with two unknowns (V and T). We can solve them simultaneously to find the values of V and T.

Please note that this calculation assumes no air resistance and a constant acceleration due to gravity.