A woman throws a ball at a vertical wall d = 5.8 m away. The ball is h = 2.9 m a

Question

A woman throws a ball at a vertical wall d = 5.8 m away. The ball is h = 2.9 m above ground when it leaves the woman's hand with an initial velocity of 17 rn/s at 45 degree. When the ball hits the wall, the horizontal component of its velocity is reversed; the vertical component remains unchanged. (Ignore any effects due to air resistance.) Where does the ball hit the ground (away from the wall) How long was the ball in the air before it hit the wall s Where did the ball hit the wall (above the ground) How long was the ball in the air after it left the wall s

Explanation / Answer

from the given data

vox = vo*cos(45) = 17*cos(45) = 12.02 m/s

voy = vo*sin(45) = 17*sin(45) = 12.02 m/s

b)

time taken for the ball to hit the wall, t1 = d/vox

= 5.8/12.02

= 0.483 s <<<<<<<<<<---------------Answer

c) Apply, H = h + voy*t1 - 0.5*g*t1^2

= 2.9 + 12.02*0.483 - 0.5*9.8*0.483^2

= 7.56 m <<<<<<<<<<---------------Answer

d) let t is the total time taken

Apply,

-h = voy*t - 0.5*g*t^2

-2.9 = 12.02*t - 0.5*9.8*t^2

4.9*t^2 - 12.02*t - 2.9 = 0

on solving the above equation we get

we get, t = 2.674 s

time taken for the ball to hit the ground after leaving the wall = 2.674 - 0.483

= 2.2 s <<<<<<<<<<---------------Answer

a) let x is the distance travelled by the ball from its initial position.

so, at x = (t - 2*t1)*vox

= (2.674 - 2*0.483)*12.02

= 20.53 m <<<<<<<<<<---------------Answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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