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Consider a falling mass(m) under gravity(9.8m/s). Initial velocity of the mass is 5 m/s upwards. Derive expressions for the velocity and the position of the mass(m) in terms of time and initial velocity/position of the mass. a) -How long will the mass take to reach the maximum height position? b)- What would be the maximum height the mass reach relative to its initial position?

Respuesta :

Answer:

a) 0.51 s

b) 1.275m

Explanation:

using equation of linear motion

v=u+gt...........................(1)

[tex]v^{2} -u^{2}=2gh[/tex]...........(2)

[tex]s=ut+\frac{1}{2} gt^2[/tex].......(3)

a) as the ball is thrown upward -ve 'g' will be acting on the body

as the body reaches to the maximum height the final velocity(v) becomes

zero so from equation (1)

0=5-9.8t

[tex]t=\frac{5}{9.8}[/tex]

t=0.51s

b) Now for maximum height calculation using equation (2)

[tex]v^{2} -u^{2}=2gh[/tex]

v=0

[tex]h=\frac{-u^2}{2g}[/tex]

[tex]h=\frac{-5^2}{2\times-9.8}[/tex]

h=1.275m

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