eet xe) 33:Velocity a e) 3.5: Projectile 3.1: Position: 1Assignments About HW 3

Question

eet xe) 33:Velocity a e) 3.5: Projectile 3.1: Position: 1Assignments About HW 3 (9 of 12) Attempt 1 of Unlimited Points Available : 1 Assignment Policy: Spring 201 Study this topic lp Time Spent 00:01:24 9 A soccer ball is kicked from ground level and misses the goal, just clearing the top of the net at the ball's highest point. The height of the top of the net is 1.49 m from the ground, and the ball is kicked from a distance of 3.71 m horizontally from the net, (a) What is the y component of the initial velocity? Q9: sbytes Regiorall Water A. Apply the equations of two- dimension kinematics to solve projectile motion problems m/s (b) What is the time it takes for the ball to reach the goal? pdf (c) What is the x component of the initial velocity? de m/s (d) At what angle above the horizontal is the ball's initial velocity? degrees

Explanation / Answer

Using the equation of motion:

v = u + at

here ,

u = ux i + uy j

v = ux i + 0 j

a = -9.8 j

(ux i ) = (ux i + uy j) + (-9.8 j)*t

So, comparing the cy-components :

0 = uy - 9.8*t

So, uy = 9.8*t

Now, using the equation of motion :

s = ut + 0.5*at^2

here , s = 3.71 i + 1.49 j

So, ( 3.71 i + 1.49 j) = (ux i + uy j)*(t) + 0.5*(-9.8 j)*t^2

So, 3.71 = ux*t ----------- (1)

and 1.49 = uy*t - 4.9*t^2

Now, uy = 9.8*t

So, 1.49 = 9.8*t^2 - 4.9*t^2

So, t = 0.551 s <----------- answer (time taken (b))

So, ux = 3.71/0.551 = 6.73 m/s <---------- answer(c)

uy = 9.8*t = 9.8*0.551 = 5.4 m/s <------- answer(a)

angle = atan(uy/ux) = atan(5.4/6.73) = 38.7 deg with respect to horizontal <------ answer (d)

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

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