A woman throws a ball at a vertical wall d = 5.0 m away.

In summary: I found the initial velocity in the x direction to be 14cos(225)=5.14 m/s and the initial velocity in the y direction to be 14sin(225)=-13.02 m/s. The equation I am trying to plug it into is V1y(SQ)=V0y(SQ)+2aΔtsorry meant Δy at the end of that equation
  • #1
cbking1306
3
0
A woman throws a ball at a vertical wall d = 3.2 m away. The ball is h = 1.6 m above ground when it leaves the woman's hand with an initial velocity of 14 m/s at 45°. 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.)

(a) Where does the ball hit the ground?
m (away from the wall)

(b) How long was the ball in the air before it hit the wall?
s

(c) Where did the ball hit the wall?
m (above the ground)

(d) How long was the ball in the air after it left the wall?
s

I found part B was .323s and part C is 4.29m. I don't know how to solve the second part.
 
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  • #2
cbking1306 said:
A woman throws a ball at a vertical wall d = 3.2 m away. The ball is h = 1.6 m above ground when it leaves the woman's hand with an initial velocity of 14 m/s at 45°. 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.)

(a) Where does the ball hit the ground?
m (away from the wall)

(b) How long was the ball in the air before it hit the wall?
s

(c) Where did the ball hit the wall?
m (above the ground)

(d) How long was the ball in the air after it left the wall?
s

I found part B was .323s and part C is 4.29m. I don't know how to solve the second part.

You should have calculated the horizontal and vertical components of the ball's velocity when it struck the wall. Use the hint at the end of the problem statement and figure out the trajectory of the ball after it bounces off the wall. (bold portion).
 
  • #3
After the ball hits the wall, you have to start a new problem with an initial velocity and height again (except going in the opposite direction).

Chet
 
  • #4
I found the initial velocity in the x direction to be 14cos(225)=5.14 m/s and the initial velocity in the y direction to be 14sin(225)=-13.02 m/s. The equation I am trying to plug it into is V1y(SQ)=V0y(SQ)+2aΔt
 
  • #5
sorry meant Δy at the end of that equation
 

Related to A woman throws a ball at a vertical wall d = 5.0 m away.

1. How do you calculate the speed of the ball?

The speed of the ball can be calculated using the formula v = d/t, where v is the speed, d is the distance traveled, and t is the time it took to travel that distance.

2. What is the initial velocity of the ball?

The initial velocity of the ball can be calculated using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity would be zero since the ball is thrown from a stationary position.

3. How long does it take for the ball to reach the wall?

The time it takes for the ball to reach the wall can be calculated using the formula t = d/v, where t is the time, d is the distance, and v is the speed. In this case, the time would be approximately 0.5 seconds.

4. How far above the ground does the ball hit the wall?

The height at which the ball hits the wall can be calculated using the formula h = ut + 1/2at^2, where h is the height, u is the initial velocity, a is the acceleration, and t is the time. In this case, the height would be approximately 1.25 meters.

5. What is the acceleration of the ball?

The acceleration of the ball can be calculated using the formula a = (v-u)/t, where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time. In this case, the acceleration would be approximately 10 m/s^2, assuming no air resistance.

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