Answer:
The velocity of the ball before it hits the ground is 381.2 m/s
Explanation:
Given;
time taken to reach the ground, t = 38.9 s
The height of fall is given by;
h = ¹/₂gt²
h = ¹/₂(9.8)(38.9)²
h = 7414.73 m
The velocity of the ball before it hits the ground is given as;
v² = u² + 2gh
where;
u is the initial velocity of the on the root = 0
v is the final velocity of the ball before it hits the ground
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 7414.73 )
v = 381.2 m/s
Therefore, the velocity of the ball before it hits the ground is 381.2 m/s
vib. motion motion of wire of guitar
circular motion revolution of earth around sun
1ml 1cm3
1m3 100cm3
volume of liquid measuring cylinder
Magnitude of acceleration
Explanation:
We know that acceleration can increase depending in the force applied on an object, any object with a greater mass will apply a greater force. F = M(a).
Answer:
x = 2
Explanation:
if it was -7 = the square root of both 2x-9 together, it would be false.
if it was square root of just 2x in the equation, the answer is:
x = 2
°°°°°°°°°
-7 = √2x - 9
-√2x = -9 + 7
√-2x = -2
√2x = 2
2x = 4
x = 2
