Question

A particle is projected vertically up from the top of a tower with velocity 10 m/s. It reaches the ground
in 5s. Find-
(a) Height of tower.
(b) Striking velocity of particle at ground
(c) Distance traversed by particle.
(d) Average speed & average velocity of particle.
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Answer 1

Answer:

height

$s = ut + \frac{1}{2} a {t}^{2} \: \:up \: direction \\ 10 \times 5 + \frac{1}{2} \times - 10 \times 25 \\ = 75m$

final velocity

before finding final velocity we have to find the time taken by object to reach ground from the maximum height

so

we use this simple equation (v = u + at) to find the time taken by the object to reach the maximum height and substract from the total time

to get the time taken by object to reach ground from the maximum height

therefore

$time \: to \: reach \: max \: height \\ v = u + at \\ 0 = 10 + 10t \\ t = 1s \\ then \: substract \: from \: total \: time \\ \: which \: is \: 5s \\ \\ 5 - 1 = 4s \\ (time \: taken \: by \: object \\ \: to \: reach \: ground \: from \: max \: height)$

then let's find the final velocity

why U is equal to zero

(that because the vertical velocity in maximum height equals to zero)

v = u+at (upward)

v= 0 + 10×4

= 40ms^-1

Total distance

look at the diagram

To find the total distance we have to add HEIGHTS OF

(A to B)+ (B to C)+ (C to D)

here we ALREADY know the height of (C to D) Which is the height of the tower 75m

but we dont know the height of( A to B )and (B to C)

The find those we have to find maximum height

${v}^{2} = {u}^{2} + 2as \: (upward) \\ 0 = { 10}^{2} + 2 \times - 10 \times s \\ 100 = 20s \\ s = 5m \\ max \: height \: = 5m$

there fore

Total distance

(A to B)+ (B to C)+ (C to D)

5 + 5 + 75

85m = total distance

$average \: speed = \frac{total \: ditance}{total \: time} \\ = \frac{85}{5} \\ = 17ms^{ - 1}$

$average \: velocity \: = \frac{total \: displacement}{total \: time} \\ = \frac - {75}{5} \\ = - 15ms^{ - 1}$

hope you can understand