t so we have to use the power rule to calculate &. Assume A in terms of A , then

Question

t so we have to use the power rule to calculate &. Assume A in terms of A , then t A,Express 8t The uncertainty in g can be ignored, but yo has an uncertainty of óyo. Express A in terms of y If the projectile is launched at a launching angle 0, with an initial velocity vo from a height of yo above the table top, how long does it take the projectile to fall on the table top? Now that your voy is no longer zero, so you must use quadratic equation to solve for t. Express the time in terms of g, yo vo and , make sure that your y-direction is fixed How far it would have traveled? Express x in terms ofg, yo, y, , and Vo lfg = 9.8 m/s, yo = 0.50 m, -300°, and vo = 10.0 m/s, how long does it take the projectile to fall on the table top? How far would it have traveled?

Explanation / Answer

t=A^(1/2)

==>dt=0.5*A^(-0.5)*dA

==>dt/t=0.5*A^(-0.5)*dA/A^(0.5)

==>dt/t=0.5*dA/A

part 2:

as A=2*y0/g

==>dA=2*d(y0)/g

==>dA/A=2*d(y0)/(g*2*y0/g)=d(y0)/y0

part 3:

initial vertical speed=v0*sin(theta)

vertical acceleration=-g

let time taken to reach the table top is t seconds

vertical distance travelled=-y0

using the formula:

displacement=initial speed*time+0.5*acceleration*time^2

==>-y0=v0*sin(theta)*t-0.5*g*t^2

==>0.5*g*t^2-v0*sin(theta)*t-y0=0

solving for t we get

t=(v0*sin(theta)+sqrt((v0*sin(theta))^2+4*0.5*g*y0))/(2*0.5*g)

part 4:

horizontal speed=v0*cos(theta)

distance travelled=horizontal speed*time

=v0*cos(theta)*(v0*sin(theta)+sqrt((v0*sin(theta))^2+4*0.5*g*y0))/(2*0.5*g)

part 5:

given :

g=9.8 m/s^2

y0=0.5 m

theta=30 degrees

v0=10 m/s

then t=1.1122 seconds

horizontal distance travelled=x=9.6316 m

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