Question

if an object is thrown vertically upwards with a speed of 49 metre per second. how long does it take to complete the upper journey. what is the maximum height achieved?

Answer 1

taking upward direction as reference(+ve)

for max height

during the upward motion of object ( from ground to max. height)

initial velocity of object, u = 49 m/s

final velocity of object, v= 0 m/s ( as object comes to standstill at max. height)

applying v^2 - u^2 = 2*g*h

[ g = acceleration due to gravity( downward) = - 9.8 m/s

h = max. height, (^ stands for power, and * for multiplication) ]

on solving, 0^2 - 49^2 = 2*(-9.8)*h

h = 122.5 m Ans

for time taken:

applying : v= u + g*t

on solving, 0 = 49 - 9.8* t

t = 5 seconds Ans

( this is the time taken by object from ground to reach to the top. Total time taken to reach the ground neglecting friction and air resistance = = 10 seconds)

Answer 2

Explanation

taking upward direction as reference(+ve)

for max height

during the upward motion of object ( from ground to max. height)

initial velocity of object, u = 49 m/s

final velocity of object, v= 0 m/s ( as object comes to standstill at max. height)

applying v^2 - u^2 = 2*g*h

[ g = acceleration due to gravity( downward) = - 9.8 m/s

h = max. height, (^ stands for power, and * for multiplication) ]

on solving, 0^2 - 49^2 = 2*(-9.8)*h

h = 122.5 m Ans

for time taken:

applying : v= u + g*t

on solving, 0 = 49 - 9.8* t

t = 5 seconds Ans

( this is the time taken by object from ground to reach to the top. Total time taken to reach the ground neglecting friction and air resistance = = 10 seconds)

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