Question

A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° above horizontal. Assume the ball encounters no air resistance, and use a Cartesian coordinate system with the origin located at the ball's initial position.

(a) Create an expression for the football’s horizontal velocity, vfx, when caught by a receiver in terms of v0, θ, g, and h.

(b) The receiver catches the football at the same height as released by the quarterback. Create an expression for the time, t f, the football is in the air in terms of v0, θ, g, and h.

(c) The receiver catches the ball at the same vertical height above the ground it was released. Calculate the horizontal distance, d in meters, between the receiver and the quarterback.

Answer 1

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = $v_{oy}$ / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

Answer 2

(a) The horizontal velocity in the projectile motion is always constant.it is the horizontal component of velocity by which the object is thrown. The expression for the football horizontal velocity will be $H = \frac{u^{2} sin^{2}\theta}{2g}$.

(b) The amount of time it takes for the body to project and land is the time of flight.

(c) The horizontal distance traveled by the ball is defined by the ball is called the range of the ball. The horizontal distance traveled will be 16.7 m.

What is a range of projectile?

The horizontal distance is covered by the body when the body is thrown at some angle is known as the range of the projectile. It is given by the formula

(a) Let the velocity at which the ball passes will be v. Resolve the velocity components into two components one is the horizontal component

$\rm{u_x = ucos\theta}$

$\rm{u_y = usin\theta}$

let the x distance is traveled in the horizontal direction so,

$y = u_y \times t$

$\rm{H = u_y t+\frac{1}{2} gt^2}$

$\rm{H = usin\theta (\frac{usin\theta}{g}) +\frac{1}{2} g(\frac{usin\theta}{g}) ^2}$

$H = \frac{u^{2} sin^{2}\theta}{g} -\frac{u^{2}sin^{2}\theta }{2g}$

$H =\frac{u^{2}sin^{2}\theta }{2g}$

This is the required relation for the horizontal velocity.

(b)

Newton's equation of motion

$\rm{v = u +gt}$

$\rm{v_y = u_y +gt}$

$\rm{v_y = 0}$

${u_y = usin\theta}$

$\rm{usin\theta = gt}$

$t= \rm{ \frac{usin\theta}{g} }$

These are the required relation for time t.

(c)

The range of the projectile is given as

$R =\frac{u^{2}sin{2}\theta }{g}$

$R =\frac{(13.5)^{2}sin64^0 }{9.81}$

R = 16.7 m

So the horizontal distance covered will be 16.7 m.

To learn more about the range of projectile refer to the link ;

https://brainly.com/question/139913