Answer:
Explanation:
Using the formula to calculate the maximum height
H = u²/2g
u is the initial velocity = 18m/s
g is the acceleration due to gravity = 9.81
H = 18²/2(9.81)
H = 324/19.62
H = 16.51m
The maximum height in which the ball reached from the ground = 85m + 16.51m = 101.5m
b) Time of fight = 2u/g
T = 2(18)/g
T = 36/9.81
T = 3.67s
It took the ball 3.67s later to reach the ground
3) To get the final velocity of the ball as it hits the ground, we need to calculate the horizontal component of the velocity,
Ux = Ucosθ
ux = 18cos 90 (angle of launch is 90 since the ball is thrown vertically upwards)
Ux = 18(0)
Ux = 0m/s
Hence the final velocity of the ball as it hits the ground is 0m/s
(a)
The tangential acceleration component of the car is simply equal to the change of the tangential speed divided by the time taken:
This rate of change is already given by the problem, 0.800 m/s^2, so the tangential acceleration of the car is
(b)
The centripetal acceleration component is given by
where
v is the tangential speed
r is the radius of the trajectory
When the speed is v = 3.00 m/s, the centripetal acceleration is (the radius is r = 10.0 m):
(c)
The centripetal acceleration and the tangential acceleration are perpendicular to each other, so the magnitude of the total acceleration can be found by using Pythagorean's theorem:
and the direction is given by:
where the angle is measured with respect to the direction of the tangential acceleration.
Answer:
Answer 1 = Google
Answer 2 = 42
Explanation:
Really, just google the words.
The 42 thing was a joke, but google 42.