a child drops a toy at rest from a point 4m above the ground at the same instant her friend throws a ball upward at 6m/s frrom a point 1m above the ground. at what distance above the ground do the ball and the toy cross paths?

d1 + d2 = 4-1 = 3 m.

(0.5g*t^2) + (Vo*t+0.5g*t^2) = 3 m.
4.9t^2 + 6t+(-4.9t^2) = 3
6t = 3
t = 0.5 s.

d1 = ho - 0.5g*t^2
d1 = 4 - 4.9*0.5^2 = 4 - 1.225=2.775 m.
Above gnd.

d2 = 1 + (6*0.5 + (-4.9)*0.5^2
d2 = 1 + (3 + (-1.225) = 1 + 1.775=2.775
m. Above gnd.

Therefore, the ball and toy cross paths
at 2.775 m. above gnd.

To find the distance above the ground where the ball and the toy cross paths, we need to determine the time it takes for each object to reach that point. Let's break down the problem step by step:

1. Firstly, let's calculate the time it takes for the ball to reach that point. We can use the following kinematic equation to find the time it takes for the ball to reach its maximum height:
h = u^2 / (2g),
where h is the height, u is the initial velocity, and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Given:
u = 6 m/s (upward)
h = 1 m

Plugging in these values into the equation, we get:
1 = 6^2 / (2 * 9.8)
1 = 36 / 19.6
19.6 = 36
Now, we can solve for time (t) by re-arranging the equation:
t = sqrt(2h / g)
t = sqrt(2 * 1 / 9.8)
t = sqrt(0.204)
t ≈ 0.452 seconds.

Therefore, it takes approximately 0.452 seconds for the ball to reach its maximum height.

2. Now, let's calculate the time it takes for the toy to fall from its initial point to the crossing point. We can use the following kinematic equation to find the time taken by the toy:
s = ut + 0.5gt^2,
where s is the distance, u is the initial velocity (0 m/s as the toy is dropped from rest), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

Given:
s = 4 m

Since the toy is dropped from rest (u = 0), the equation simplifies to:
s = 0.5gt^2

Plugging in the values, we can solve for time (t):
4 = 0.5 * 9.8 * t^2
4 = 4.9 * t^2
t^2 = 4/4.9
t^2 ≈ 0.8163
t ≈ sqrt(0.8163)
t ≈ 0.904 seconds.

Therefore, it takes approximately 0.904 seconds for the toy to fall to the crossing point.

3. Finally, to find the distance above the ground where the ball and the toy cross paths, we need to determine the distance traveled by the ball in 0.452 seconds. We can use the equation of motion:
s = ut + 0.5gt^2,
where s is the distance, u is the initial velocity (6 m/s upward), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time (0.452 seconds).

Plugging in these values, we get:
s = 6 * 0.452 + 0.5 * 9.8 * 0.452^2
s = 2.712 + 0.5 * 9.8 * 0.204
s = 2.712 + 1.004
s ≈ 3.716 meters.

Therefore, the ball and the toy cross paths approximately 3.716 meters above the ground.