Question

A particle is projected vertically up from the top of a tower with velocity 10m/s . It reaches ground in5s. Find height of tower and distan
ce traversed by particle???

Answer 1

Answer:

The height or distance traveled by the particle is 83.5 m.

Explanation:

Given that,

Velocity v = 10 m/s

Time t = 5 sec

We need to calculate the distance from the top of a tower

Using third equation of motion

$v^2=u^2+2gs$

Where, v= final velocity

u = initial velocity

g = acceleration due to gravity

s = distance

Put the value in the equation

$0=10^2-2\times9.8\times s'$

$s'=\dfrac{100}{2\times9.8}$

$s' =5.1\ m$

For time,

The time at maximum height,

Using first equation of motion

$v = u+gt$

$0=10-9.8\times t'$

$t'=\dfrac{10}{9.8}$

$t' =1.0\ sec$

The maximum time is

$t''=t-t'$

$t''=5-1=4\ sec$

Now, the particle reached at the ground from the maximum time in 4 sec.

For the distance traveled by the particle is

Using second equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value in the equation

$s''=0+\dfrac{1}{2}\times9.8\times4\times4$

$s''=78.4\ m$

The total distance is

$s = s'+s''$

$s=5.1+78.4=83.5\ m$

Hence, The height or distance traveled by the particle is 83.5 m.