Answer:
a= -0.86 m/s²
The negative sign shows that ball down the ground or moving down
Explanation:
Vf² - Vo² = 2gS
where
Vf = velocity of clay as it hits the ground
Vo = initial velocity of clay = 0
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
S = distance travelled by clay = 15 m
Substituting appropriate values,
Vf² - 0 = 2(9.8)(15)
Vf = 17.15 m/sec.
Formula to use is,
V - Vf = aT
where
V = velocity of clay when it stops = 0
Vf = 17.15 m/sec (as determined above)
a = acceleration
T = 20 ms
Put the values to find acceleration
a=(V-Vf)/T
a=(0-17.15)/20
a= -0.86 m/s²
The negative sign shows that ball down the ground
Answer:
Resultant force, R = 10 N
Explanation:
It is given that,
Force acting along +x direction, 
Force acting along +y direction, 
Both the forces are acting on a point object located at the origin. Let the resultant force of the object is given by R. So,
Here 
R = 10 N
So, the resultant force on the object is 10 N. Hence, this is the required solution.
According to another source this is what I got
<span>0.735 J ( Ep-potential energy, m-mass,g-gravitational acceleration = 9.81m/s², h-height; Ep = m * g * h; Ep = 0.0300 kg * 9.81 m/s² * 2.5 m )
</span>Hope it helps
Answer:
B
Explanation:
From Newton's law of motion, we have:
V^2 = U^2 + 2gH
Where V and U are final and initial velocity respectively.
H is the height.
For the object to have a sustain a maximum height it means the final velocity of the object is zero.
By computing the height of the object sustain by A, we have:
0^2 = 2^2 -2×10×H
0= 4 -20H
4 = 20H;
H= 0.2m
For object B we have;
0^2 = 1^2 -2×10×H
0 = 1 -20H
H = 1/20= 0.05m
From computing the height sustain by both objects, we see object B is projected at a shorter height into atmosphere than A.
Hence object B will return to the ground first.
