Question

1. A projectile is launched horizontally from a table, with an initial velocity of 5.0 m/s. When the projectile lands on the ground, it has traveled 11 m forward. What is the height of the table?

2. A projectile is launched horizontally from 2.6 m off the ground. The projectile lands on the ground 15 m forward from where it was launched. Find the projectile's initial velocity.

Answer 1

1. The height of the table is 2.352 meters.

2. The initial horizontal velocity of the projectile is 10 meters per second.

Solution 1:

Given:

Initial horizontal velocity (v_x) = 5.0 m/s

Horizontal distance traveled (d) = 11 m

Unknown:

Height of the table (h)

Assumptions:

Acceleration due to gravity (g) = 9.8 m/s^2

Air resistance is negligible

Equations:

Horizontal displacement: d = v_x * t

Vertical displacement: h = 0.5 * g * t^2

From the horizontal displacement equation, we can solve for time (t):

t = d / v_x = 11 m / 5.0 m/s = 2.2 seconds

Substitute the time (t) into the vertical displacement equation:

h = 0.5 * 9.8 m/s^2 * (2.2 seconds)^2 = 2.352 m

Solution 2:

Given:

Height of the table (h) = 2.6 m

Horizontal distance traveled (d) = 15 m

Unknown:

Initial horizontal velocity (v_x)

Assumptions:

Acceleration due to gravity (g) = 9.8 m/s^2

Air resistance is negligible

Horizontal displacement: d = v_x * t

Vertical displacement: h = 0.5 * g * t^2

From the vertical displacement equation, we can solve for time (t):

t = sqrt(2h / g) = sqrt(2 * 2.6 m / 9.8 m/s^2) = 1.5 seconds

Substitute the time (t) into the horizontal displacement equation:

v_x = d / t = 15 m / 1.5 seconds = 10 m/s