Question

The projectile is again launched from the same position, but with the cart traveling to the right with a speed v1 relative to the ground, as shown below (third image). The projectile again leaves the cart with speed vo relative to the cart at an angle θ above the horizontal, and the projectile lands at point Q, which is a horizontal distance D from the launching point. Express your answer in terms of vo, θ, and physical constants, as appropriate.(3) Give a physical reason why the projectile lands at point Q, which is not as far from the launch position as point P is, andexplain how that physical reason affects the flight of the projectile.(4) Derive an expression for v1. Express your answer in terms of vo, θ, D, and physical constants, as appropriate.After the launch, the cart’s speed is v2. Beginning at time t = 0, the cart experiences a braking force of F = -bv, where b is a positive constant with units of kg/s and v is the speed of the cart. Express your answers to the following in terms of m, b, v2, and physical constants, as appropriate.

Answer 1

The projectile lands at point Q, which is closer to the launch position than point P, due to the cart's initial forward velocity v1 altering the horizontal motion of the projectile. This forward motion reduces the relative horizontal distance traveled by the projectile before landing.

The expression for v1 is derived as follows:

`v1=(D)/(t) +v2`.

where
`t` is the time taken for the projectile to land at point Q.

Step-by-step explanation:

The projectile's horizontal motion is affected by the initial velocity of the cart, resulting in a shorter horizontal distance traveled. As the cart moves forward with a velocity
`v1` relative to the ground, the projectile launched from it inherits this horizontal velocity component. Consequently, the projectile's initial horizontal velocity is the sum of the cart's velocity
`v1` and the projectile's initial horizontal component. This alters the horizontal distance the projectile covers before landing, leading to a shorter range (point Q) compared to when launched from a stationary cart.

To determine
`v1`, consider the horizontal distance
`D` traveled by the projectile and the time
`t` it takes to reach point Q. The cart's final velocity
`v2` is also a factor in the calculation. Using the kinematic equation for constant acceleration in the horizontal direction, the expression for
`v1` is
`v1=(D)/(t) +v2` This equation incorporates the distance
`D` traveled by the projectile relative to time
`t` and the cart's final velocity
`v2` after braking, providing the value for
`v1\\` in terms of the given variables and physical constants.

Answer 2

The solution to the projectile motion problem involves separating the motion into horizontal and vertical components and using kinematic equations to determine the time of flight, range, and initial velocity of the cart. The projectile's trajectory is affected by both its initial velocity relative to the cart and the cart's velocity relative to the ground.

Step-by-step explanation:

The key to solving projectile motion problems is to treat the vertical and horizontal components separately. Given an initial speed vo and launch angle θ, we can determine the horizontal and vertical components of the velocity. The horizontal component, which determines how far the projectile travels, will be vo cos(θ) while the vertical component, which determines the flight time of the projectile, is vo sin(θ). When the projectile lands at point Q a distance D from the launch point, the initial velocity of the cart v1 additively combines with the horizontal component of the projectile's launch speed.

To determine v1, we can use the formula D = (v1 + vo cos(θ)) * t, where t is the time the projectile is in the air, which is determined by the vertical motion. Applying the kinematic equations to the vertical motion, we can find the time t based on vo sin(θ) and the acceleration due to gravity, g. When the cart experiences the braking force F = -bv, we can use the relationship between force, mass m, and acceleration to find the deceleration of the cart and subsequently, how its velocity changes over time after the projectile is launched.