Question

A ball I thrown vertically upwards with a speed of 19.6m/s From top of a tower returns to the earth in 6 s. What I height of the tower. [take g= 9.8m/s^2]​

Answer 1

$\bold \pink{initial \: velocity = u = 19.6m {s}^{ - 2} }$

$\bold \pink{time \: taken = t = 6 s}$

The magnitude of displacement is the height of the tower . Final position is in downward direction with respect to initial position (top of tower)

$\bold \blue{s = - h}$

Acceleration is in downward direction and is negative.

$\tt \green{a = g = - 9.8m {s}^{} }$

$\tt \: from \: second \: equation \: of \: motion \: we \: have :$

$\tt {s = u + \frac{1}{2}a {t}^{2} }$

$\tt - h= 19.6 \times 6 - \bigg( \frac{1}{2} \times 9.8 \times {6}^{2} \bigg)$

$\tt = 58.8m {s}^{ - 1}$

Height = 58.8m/s

Answer 2

$\bf{\green{\underline{Given:}}}$

Initial velocity ,u = 19.6m/s

Final velocity , v = 0

Time taken , t = 6 sec

acceleration due to gravity ,g = 9.8m/s^2

(when a body is thrown vertically upwards it's rate of change of velocity goes on decreasing ,hence the value of g will be negative )

Acceleration due to gravity,g = -9.8 m/s^2

$\bf{\red{\underline{To\;Find}}}$

Height of the tower

$\bf{\pink{\underline{Solution:}}}$

Using Second equation of motion :

$\boxed{\green{s=ut+ \frac{1}{2}g{t}^{2}}}$

where,

s is distance covered

u is initial velocity

a is acceleration produced

t is time taken

${s = ut + \frac{1}{2}a {t}^{2}}$

${s = 19.6 \times 6 + \frac{1}{2} \times ( - 9.8) \times {6}^{2}}$

${s = 117.6 + \frac{1}{2} \times ( - 9.8) \times 36}$

$\sf{s=117.6+\frac{1}{\cancel 2} \times (-9.8)\times \cancel 36}$

$\sf{s=117.6 + (-9.8)\times 18}$

$\sf{s=117.6-9.8\times 18}$

$\sf{s=58.8m}$