Question

A ball is dropped from the top of a straight cliff. the ball reaches the ground 4second later. how high is the cliff? what is the final velocity of the ball when it just reaches the ground? (take g= 10 m s^-2​

Answer 1

V = 40 m/sec

S = 80 m

Given:

Time(t) = 4 sec

To Find:

Velocity(v)

Height Of Straight Cliff (h)

How To get Solution:

$\sf \: Use \: formula \\ \sf a) \: v = u + at \: \ \\ \sf \: b)\: s \: = \: ut \: + \frac{1}{2} a{t}^{2}$

[here, u initial velocity, s is height and a is acceleration due to gravity]

u = 0

Now,

$v \: = 0 + 10 \times 4 \\ v \: \sf = 40 \: ms {}^{ - 1}$

Again,

$\sf \: s \: = \: ut \: + \frac{1}{2} a{t}^{2} \\ \\ = 0 \times 4 + \frac{1}{2} \times 10 \times 4 {}^{2} \\ \\ = 5 \times 16 \\ \\ \sf = 80 \: m$

If, I ma rihgt thne teh anwesr si 80 m and 40 m/s

Answer 2

Given :-

Time = 4 sec

G = 10 m/s

u = 0 m/s

To Find :-

Height

Final velocity

Solution :-

We know that

$\large \sf v = u + at$

=> v = 0 + (10)(4)

=> v = 0 + 40

=> v = 40 m/s

Now

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

=> s = 0 × 4 + ½ × 10 × 4²

=> s = 0 + 5 × 16

=> s = 0 + 80

=> s = 80 m