The position function of a projectile is given by: r(t) = (5.0 t + 6.0 t 2 ) m i

ID: 2111657 • Letter: T

Question

The position function of a projectile is given by:

r(t) = (5.0 t + 6.0 t2) m i + (30 â€“ t3) m j

(a) From what height above the ground was the projectile launched?

(b) What is the displacement of the particle at t = 2.0 s?

(d) What is the horizontal displacement of the projectile when it lands on the ground?

(e) What is the time taken by the particle to reach the maximum height?

(f) What is the velocity of the projectile (in magnitude angle form) when it lands on the ground?

a)The height can by calculated by putting t=0 in the projectile's y coordinate.

height=30-*0=30m

b)r(2)=(5*2+6*4)i+(30-8)j=34m i -22m j

c)To land on the ground, y component=0.

30-t3=0

t=3.107s

d)x(3.107 )=5*3.107+6*3.107^2 = 73.455694 m

e)for maximum height,

dy/dt=-3t2=0 i.e. the particle was at maximu height at t=0.

f)v=dr/dt=(5+12t)i-3t2j

v(3.107)=42.284i-28.960347j

v(1.33)=51.25m/s at an angle 34.407 degrees below horizontal.

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