The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²
The direction of the acceleration of the ball is downwards
The given parameters
initial velocity of the ball, u = 0
height above the ground, h = 2.2 m
time of motion of the ball, t = 96 ms = 0.096 s
The magnitude of the acceleration of the ball while coming to rest is calculated as;
let the downwards direction of the acceleration be positive
The direction of the acceleration of the ball is downwards
Learn more here: brainly.com/question/15407740
Answer:
2.605m
Explanation:
Using the formula for calculating Range (distance travelled in horizontal direction)
Range R = U√2H/g
U is the speed = 4.8m/s
H is the maximum height = ?
g is the acc due to gravity = 9.8m/s²
R = 3.5m
Substitute into the formula and get H
3.5 = 4.8√2H/9.8
3.5/4.8 = √2H/9.8
0.7292 = √2H/9.8
square both sides
0.7292² = 2H/9.8
2H = 0.7292² * 9.8
2H = 5.21
H = 5.21/2
H = 2.605m
Hence the height of the ball from the ground is 2.605m
Answer:
White hole is an impossible object in universe. ... This means that in a hypothetical universe where there is a black and a white hole, in a short time after their first interaction the white hole will become another black hole so that the system will end up with two black holes.
Answer: A projectile is any object in which the only force is gravity
Explanation: Equations on how to calculate projectile velocity is stated below:
The initial velocity Vo being a vector quantity, has two componentsVox and Voy
V0x = V0 cos(θ)
V0y = V0 sin(θ)
The acceleration A is a also a vector with two components Axand Ay given
Ax = 0 and Ay = - g = - 9.8 m/s2
Along the x axis the acceleration is equal to 0 and therefore the velocity Vx is constant
Vx = Vocos(θ)
Along the y axis, the acceleration is uniform and equal to - g and the velocity at time t is g
Vy = Vo sin(θ) - g t
Along the x axis the velocity Vx is constant and therefore the component x of the displacement is
x = Vocos(θ) t
Along the y axis, the motion is of uniform acceleration and the y component of the displacement is
y = Vo sin(θ) t - (1/2) g t2
