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
Answer:
9ms^2
Explanation:
since ,Force=mass*acceleration
then, acceleration=force/mass
and, Force=90N
Mass=10pound
therefore, acceleration=90/10
=9ms^2
Answer:
35 kg
Explanation:
From the question,
Momentum (I) = mass (m) × velocity (v)
I = m×v................... Equation 1
Where m = mass, v = velocity
make m the subject of the equation
m = I/v.................... Equation 2
Given: I = 140 kgm/s, v = 4 m/s
Substitute these values into equation 2
m = 140/4
m = 35 kg
Hence the mass of the dart is 35 kg
Thank you for posting your question here at brainly. A mass of m moves with 2V towards in the opposite direction of a mass, 4m moving at a speed of V, the speed of m was 2/5V and the mass of 4m was 7.5V. I hope it helps.
