The height at time t is given by
h(t) = -4.91t² + 34.3t + 1
When the ball reaches maximum height, its derivative, h'(t) = 0.
That is,
-2(4.91)t+34.3 = 0
-9.82t + 34.3 = 0
t = 3.4929 s
Note that h''(t) = -9.82 (negative) which confirms that h will be maximum.
The maximum height is
hmax = -4.91(3.4929)² + 34.3(3.4929) + 1
= 60.903 m
Answer:
The ball attains maximum height in 3.5 s (nearest tenth).
The ball attains a maximum height of 60.9 m (nearest tenth)
Answer: D. 0.29 m
Explanation:
We will use the following equations to describe the leap of the cat:
(1)
(2)
Where:
is the height of the cat
is the cat's initial velocity
is the acceleration due gravity
is the time
is the y-component of the velocity
Now the cat will have its maximum height 
when 
. So equation (2) is rewritten as:
(3)
Finding :
(4)
(5)
(6)
Substituting (6) in (1):
(7)
Finally:
(8)
Ken jenr ewjk hfjek kwe hwlr herw hriehr jkwehrt
C^2=a^2+b^2
c^2=45^2+45^2
c^2=4050
c=63.64
c=64
The answer is D.
