Question

A stone is thrown vertically upward with a speed of 28 m/s. (a) Find the maximum height reached by the stone, (b) Find its velocity one second before it reaches the maximum height, (c) Does the answer of part (b) change if the initial speed is more than 28 m/s such as 40 m/s or 80 m/s. Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

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Answer 2

Solution :

Initial speed=u=28m/s
Final velocity at maximum height=v= 0 m/s
a=g=10m/s2

A) maximum height attained by stone=u²/2g=28x28/2x10=39.2m

B) time=t=u/g=28/10=2.8 sec
t1=2.85-1=1.85
V=u+at
=28 -(10)(1.85)=28-18.13=9.87 m/s
Hence the velocity is 0.87 m/s

c) No it will not change .
Because after 1 sec final velocity becomes zero for any initial velocity..