When air resistance is negligible, a projectile near the surface of the earth is

Question

When air resistance is negligible, a projectile near the surface of the earth is subjected only to the force of gravity. In that case, Newton's second law tells us that the projectile's acceleration is constant and downward. Denote by (x, y) the position of the projectile, such that y is the vertical distance from the ground, and x is the distance along the ground in the direction of motion. Then the projectile's acceleration (a_x, a_y) has the following components: a_x = d^2x/dt^2 = 0, a_y = d^2y/dt^2 = -g, Where t is time, and g is the local acceleration due to gravity near the earth's surface (in SI units, g almostequalto 9.8 m/s^2). (a) Starting from (2), integrate to find the projectile's velocity components v_x(t) = dx/dt and v_y(t) = dy/dt as functions of time t, subject to the initial conditions v_x(0) = v_x, 0 and v_y(0) = v_y, 0. Leave your answer in terms of g. (b) Integrate again to find the projectile's position components x(t) and y(t) as functions of time t, subject to the initial conditions x(0) = x_0 and y(0) = y_0. Leave your answer in terms of g. (c) From your expression for x(t, ), solve for t, (x) (that is, t as a function of x). Then substitute t(x) into your expression for y(t) to find the projectile's trajectory y(x) (that is, y as a function of x). [Note: You do not need to expand your expression for y(x).] What is the shape of the projectile's trajectory? (d) Suppose you throw a baseball of mass 0.145 kg from an initial height of y_0 = 1 m above the ground at a speed of 40.2 m/s (90 mph) at an angle of 20 degree from the horizontal. (i) How long does it take for the ball to hit the ground? (ii) How far does the ball travel horizontally during that time? (iii) What is the maximum height of the ball above the ground?

Explanation / Answer

a. given ax = 0

so ax = d(vx/dt)

so, vx = c [ where c is a constant]

but vx(0) = v(x,0)

so, vx = (vx,0) = constant

similiarly

ay = dvy/dt = -g

dvy = -g*dt

vy = -gt + k

v(y,0) = k

so, vy = v(y,0) - gt

b. dx/dt = vx = v(x,0)

integrating

x = v(x,0)*t + xo

similiarly

dy/dt = vy = v(y,0) - gt

y = v(y,0)t - 0.5gt^2 + yo

c. x = v(x,0)*t + xo

t(x) = (x - xo)/v(x,0)

y = v(y,0)t - 0.5gt^2 + yo

substituting

y - yo= v(y,0)((x - xo)/v(x,0)) - 0.5g((x - xo)/v(x,0))^2

the trajectory is of the shape of a parabola

d. m = 0.145 kg

yo = 1 m

v(x,0) = 40.2cos(20) = 37.775 m/s

v(y,0) = 40.2sin(20) = 13.749 m/s

xo = 0

time taken to hit the ground = t

using the equaiton

h = ut - 0.5gt^2

-1 = 13.749*t - 0.5*g*t^2

4.905t^2 - 13.749t - 1 = 0

a. t = 2.873 s

b. range, r = v(x,0)t = 108.5275 m

c. maximum height = H

-2*g*(H-1) = -v(y,0)^2

2*9.81*(H-1) = 13.749^2

H = 10.634 m

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