Answer:
6.5 m above the floor and 5 m above Christine's hand when it reaches the maximum height.
Explanation:
Let g = 10 m/s2 be the gravitational deceleration that affects the ball vertical motion so it comes to the maximum height at 0 speed. We can use the following equation of motion to find out the distance traveled by the ball from where it's thrown:
[tex]v^2 - v_0^2 = 2g\Delta s[/tex]
where v = 0 m/s is the final velocity of the ball when it reaches maximum level, [tex]v_0[/tex] = 10m/s is the initial velocity of the ball when it starts, g = -10 m/s2 is the deceleration, and [tex]\Delta s[/tex] is the distance traveled, which we care looking for:
[tex]0^2 - 10^2 = -2*(-10)\Delta s[/tex]
[tex]\Delta s = 100 / (2 * 10) = 5 m[/tex]
So the ball is 5 m above Christine' hands when it reaches maximum height, and since the hand is 1.5 m above the floor, the ball is 5 + 1.5 = 6.5 m above the floor when it reaches maximum height.
Answer:
6.602m above the floor
5.102m above christine's hand.
Explanation:
This problem involves the concept of motion under free fall.
At maximum height the velocity is 0m/s
V = u -gt
u = 10m/s and g =9.8m/s²
t is the time taken to reach maximum height and is unknown.
On substitution
0 = 10 - 9.8t
9.8t = 10
t = 10 / 9.8
t = 1.02s
H = h + ut -1/2gt²
H is the height
h is the initial height = 1.5m
H = 1.5 + 10×1.02 - 1/2 ×9.8×1.02² = 6.602m
The distance the ball travels above christine's hand is 6.602 - 1.50m = 5.502m
