Question

A stone is thrown vertically upwards with a velocity of 19.6 m/s
a) Find the velocity and acceleration of the stone at the highest point of its motion
b) How high will a particle rise?
c) What time will it take to reach the highest point and come back?

Answer 1

Answer:

☆Concept:

》whenever object is thrown above the surface,it will face downward acceleration due to gravity,i.e,9.8m$\ s^{-2}$.

》so it will create retardation in its motion,after some time it will become momentarily at rest and start to fall down in dirn of gravity.

》so at maximum height achieved it velocity will be 0.

》one thing here is to note that,during the motion from ground to up and again to ground its net displacement will be zero in the y-direction ,

as its initial ad final position are same.

☆Given:

》initial velocity(u) =19.6m/s

☆Formula applied:

• $\boxed{\bf{s=ut+ \dfrac{1}{2}at^{2}}}$

• $\boxed{\bf{v=u+at}}$

☆Solution:

a)velocity and acceleration at highest point?

$\longrightarrow{\sf{velocity =0}}$

$\longrightarrow{\sf{acceleration=g=9.8ms^{-2}}}$

b)maximum height?

$\longrightarrow{\sf{v=u+at}}$

$\longrightarrow{\sf{0=19.6-9.8t}}$

$\longrightarrow{\sf{t= \dfrac{19.6}{9.8}}}$

$\longrightarrow{\bf{time=2secs}}$

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$\longrightarrow{\sf{s=ut+\dfrac{1}{2}at^{2}}}$

$\longrightarrow{\sf{s= 19.6×2+\dfrac{1}{2}9.8(2)^{2}}}$

$\longrightarrow{\sf{s=58.8}}$

$\longrightarrow{\bf{H_{max}=58.8meters}}$

c)total time for up and down?

》we have already find the time to reach up =2secs

$\longrightarrow{\sf{s=ut+\dfrac{1}{2}at^{2}}}$

••here,s will be net displacement from up to down.

$\longrightarrow{\sf{0=19.6t +\dfrac{1}{2}(-9.8)t^{2}}}$

$\longrightarrow{\sf{19.6t = 4.9t^{2}}}$

$\longrightarrow{\bf{total\;time=4secs }}$

》$\ t_{up}=t_{down}= 2secs$

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♧hope it helps!♧