Question

a bullet is fired at an angle theta above the horizontal with an initial velocity: 800 m/s from the top of an 80 m tower. What value of theta will give the maximum horizontal range?

Answer 1

40.8 degrees

Step-by-step explanation:

To find the angle theta that will give the maximum horizontal range, we can use the following steps:

1) Break the initial velocity into its horizontal and vertical components. The horizontal component of the velocity remains constant throughout the motion, while the vertical component changes due to the effect of gravity.

The initial horizontal velocity of the bullet is:

Vx = V0 cos(theta)

where V0 is the initial velocity of the bullet (800 m/s) and theta is the angle above the horizontal.

The initial vertical velocity of the bullet is:

Vy = V0 sin(theta)

2) Determine the time of flight of the bullet. The time of flight is the time it takes for the bullet to reach the ground.

The vertical distance traveled by the bullet can be calculated using the equation:

y = V0 sin(theta)t - (1/2)gt^2

where y is the vertical distance traveled by the bullet, g is the acceleration due to gravity (9.8 m/s^2), and t is the time of flight.

At the maximum horizontal range, the vertical distance traveled by the bullet is equal to the height of the tower (80 m). Therefore, we can set y = 80 m and solve for t:

80 = V0 sin(theta)t - (1/2)gt^2

We can use the quadratic formula to solve for t:

t = [V0 sin(theta) +/- sqrt((V0 sin(theta))^2 + 2gy)]/g

The positive solution gives the time of flight of the bullet. We can substitute this value of t into the equation for the horizontal distance traveled by the bullet to find the maximum horizontal range:

R = V0 cos(theta)t

3) Differentiate the horizontal range with respect to theta and set the derivative equal to zero to find the angle that maximizes the horizontal range.

dR/dtheta = -V0sin(theta)t + V0cos(theta)(dt/dtheta)

Setting dR/dtheta equal to zero and solving for theta gives:

tan(theta) = g*t/V0

Substituting the expression for t from above into this equation gives:

tan(theta) = (2V0^2/g)*sin(theta)*cos(theta)

This equation can be solved numerically to find the value of theta that maximizes the horizontal range. In this case, we can use a graphing calculator or computer software to solve for theta, which is approximately 40.8 degrees.

Therefore, the value of theta that will give the maximum horizontal range is approximately 40.8 degrees.

Answer 2

90 degrees (or pi/2 radians)

Step-by-step explanation:

To find the value of theta that will give the maximum horizontal range, we need to consider the projectile motion of the bullet. When a projectile is fired at an angle above the horizontal, it follows a parabolic trajectory. The key to maximizing the horizontal range is to maximize the time of flight of the projectile. The time of flight can be determined using the vertical component of the initial velocity and the acceleration due to gravity. First, let's break down the initial velocity of 800 m/s into its vertical and horizontal components. The vertical component can be found using the sine function, and the horizontal component can be found using the cosine function. Vertical component: V_y = 800 * sin(theta) Horizontal component: V_x = 800 * cos(theta) The time of flight, t, can be determined by dividing the vertical displacement by the vertical component of velocity: Vertical displacement = 80 m (height of the tower) t = (vertical displacement) / (V_y) Now, we can determine the horizontal range, R, using the horizontal component of velocity and the time of flight: R = V_x * t To maximize the horizontal range, we need to maximize the time of flight, t. Since the vertical component of velocity depends on theta, we can maximize t by maximizing V_y. The maximum value of sin(theta) is 1, which occurs when theta = 90 degrees. Therefore, to maximize the horizontal range, theta should be 90 degrees (or pi/2 radians).