Question

A toy rocket is launched with an initial velocity of 14.0 m/s in the horizontal direction from the roof of a 34.0-m -tall building. the rocket's engine produces a horizontal acceleration of ( 1.60 m/s³)t , in the same direction as the initial velocity, but in the vertical direction the acceleration is g , downward. air resistance can be horizontal distance does the rocket travel before reaching the ground?

Answer 1

To calculate the horizontal distance the toy rocket will travel before hitting the ground, separate the motion into vertical and horizontal components. It will take approximately 2.65 seconds for the rocket to fall 34.0 meters. During this time, it will travel approximately 42.7 meters horizontally due to its initial velocity and acceleration.

Step-by-step explanation:

Calculating the Horizontal Distance of the Toy Rocket

To find the horizontal distance the rocket travels before reaching the ground, we need to consider the vertical and horizontal motions separately, as they are independent of each other. Given that the toy rocket is launched horizontally from a 34.0-meter-tall building at an initial velocity of 14.0 m/s, and experiences a horizontal acceleration of (1.60 m/s³)t, we can calculate the time it takes to hit the ground due to gravity.

First, determine the time taken for the rocket to fall 34.0 meters using the equation for vertical motion:

• s = ut + 0.5gt²
• 34.0 = 0 + 0.5(9.8)t²
• t = √(34.0 / 4.9)
• t = 2.65 seconds (approximately)

Next, calculate the horizontal distance using the constant initial velocity and the acceleration:

• x = ut + 0.5at²
• x = (14.0 m/s)(2.65 s) + 0.5(1.60 m/s³)(2.65 s)²
• x = 37.1 m + 5.6 m
• x = 42.7 meters (approximately)

The toy rocket will travel approximately 42.7 meters horizontally before it hits the ground.

Answer 2

The toy rocket travels a horizontal distance of approximately 36.7 meters before reaching the ground.

Step-by-step explanation:

To find the horizontal distance the toy rocket travels before reaching the ground, we first need to determine the time it takes for the rocket to hit the ground. Since the rocket's acceleration in the vertical direction is the acceleration due to gravity (9.8 m/s²), we can use the equation:

Δy = v0t + 0.5at²

where Δy is the vertical displacement (negative since the rocket is moving downward), v0 is the initial vertical velocity (0 m/s), a is the acceleration (-9.8 m/s²), and t is the time.

Solving for t gives:

t = √(2Δy/a)

Substituting in the values, we have:

t = √(2(-34)/(-9.8)) = √(68/9.8) ≈ 2.627 s

Now, we can find the horizontal distance using the equation:

Δx = v0xt + 0.5at²

where Δx is the horizontal displacement, v0x is the initial horizontal velocity (14.0 m/s), and t is the time (2.627 s).

Substituting in the values, we have:

Δx = (14.0 m/s)(2.627 s) = 36.7 m

Therefore, the toy rocket travels a horizontal distance of approximately 36.7 meters before reaching the ground.