A ball of mass 184 g is thrown with an initial velocity of 25 m/s at an angle o

Question

A ball of mass 184 g is thrown with an initial velocity of 25 m/s at an angle o with the horizontal direction. Ignore air resistance. What is the ball after 0.6 s? (Do t momentum of the his problem by finding the components of the momentum first r from and then constructing the magnitude and direction of the momentum vecto the components. Find the magnitude in kg m/s and the direction in degrees counterclockwise from the +x-axis. Assume the +x-ax is points horizontally to the right.) , (25 m/s)v 30°

Explanation / Answer

given m = 184 g
v = 25 m/s
theta = 30 deg

after t = 0.6 s
let the vertical and horixontal veloctiy componetns be vx, vy
then
vx = vcos(theta)
vy = vsin(theta)*t - 0.5gt^2
hence
vy = 25*sin(30)*0.6 - 0.5*g*0.6^2= 5.7342 m/s
vx = 21.65063509 m/s

hence v' = sqroot(vx^2 + vy^2) = 22.397121458 m/s
momentum = mv' = 4.12107034 kg m/s
direction , theta = arctan(vy/vx) = 14.8342687 deg CCW to the +x axis rtunning to right

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.