Answer: Ax=(Vx-Vox)/(T)
Vx=Vox+Ax*T
Solving for Ax in terms of Vx, Vox, T
Vx-Vox=Ax*t
Ax=(Vx-Vox)/(T)
This is saying the acceleration in the x-direction can be found by taking the difference between the finial and initial Velocity in x-direction and dividing it by the Total Time.
Any questions please feel free to ask. Thanks
U=10 m/s
v=30 m/s
t=6 sec
therefore, a=(v-u)/t
=(30-10)/6
=(10/3) ms^-2
now, displacement=ut+0.5*a*t^2
=60+ 0.5*(10/3)*36
=120 m
And you can solve it in another way:
v^2=u^2+2as
or, s=(v^2-u^2)/2a
=(900-100)/6.6666666.......
=120 m
Answer: 13.2 seconds.
Explanation: using equation of motion; S= ut +1/2at² where u = initial velocity=0
S= distance travelled
a = acceleration due gravity
t= time.
1 foot = 0.305m so,
S= 2860 feet =872.3m
S= ut+1/2 at²
872.3 = 0×t + 1/2×10 × t²
872.3 =0 + 5t²
T²= 872.3/5
T²= 174.46
Take the square root of T we then have;
t = 13.2 seconds to one decimal place.
Answer:
v = 29.4m/s
Explanation:
Since the ball is dropped at rest,
u = 0m/s
a = 9.81m/s²
Using
v = u + at
After 3 seconds,
v = 0 + (9.81)(3)
v = 29.4m/s
