Question

a stone thrown upward from the top of tower 85 m high, reaches ground in 5 sec.
find
the greatest hight above ground
velocity with which it reaches the ground
time taken to reach maximum hight.
take g=10m/s-2​

Answer 1

$\huge\mathfrak{\underline{Solution}}$

$\sf{Net\:displacement}$

$\longrightarrow$$s = - 85m$

$\longrightarrow$$t = 5s$

$\longrightarrow$$g = - 10m {s}^{ - 2}$

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

$\longrightarrow$$- 85 = u \times 5 - \frac{1}{2} \times 10 \times 25$

$\longrightarrow$$u = 8m {s}^{ - 1}$

1)Height covered in upward motion is given by -

$\longrightarrow$${v}^{2} = {u}^{2} + 2gs$

$\longrightarrow$$0 = {8}^{2} - 2( - 10)s$

$\longrightarrow$$s = 3.2m$

The greatest height above the ground

$\longrightarrow$$85 + 3.2 m$

$\longrightarrow$$88.2m$

2)For the displacement motion-

$\longrightarrow$$u = 0$

$\longrightarrow$$g = 10m {s}^{ - 2}$

$\longrightarrow$$s = 88.2m$

$\longrightarrow$${v}^{2} = {u}^{2} + 2gs$

$\longrightarrow$$0 + 2 \times 10 \times 88.2$

$\longrightarrow$$1764$

3)Let t be time taken to reach the maximum height

For upward motion

$\longrightarrow$$u = 8m {s}^{ - 1}$

$\longrightarrow$$v = 0$

$\longrightarrow$$g = - 10m {s}^{ - 1}$

$\longrightarrow$$t = \frac{v - u}{g}$

$\longrightarrow$$t = \frac{ - 8}{ - 10}$

$\longrightarrow$$t = 0.8s$

Answer 2

displacement

⟶ s = - 85ms=−85m

⟶ t = 5st=5s

⟶ g = - 10m {s}^{ - 2}

g=−10ms

−2

⟶ s = ut + \frac{1}{2} g {t}^{2}s=ut+

2

1

gt

2

\longrightarrow⟶ - 85 = u \times 5 - \frac{1}{2} \times 10 \times 25−85=u×5−

2

1

×10×25

\longrightarrow⟶ u = 8m {s}^{ - 1}u=8ms

−1

1)Height covered in upward motion is given by -

\longrightarrow⟶ {v}^{2} = {u}^{2} + 2gsv

2

=u

2

+2gs

\longrightarrow⟶ 0 = {8}^{2} - 2( - 10)s0=8

2

−2(−10)s

\longrightarrow⟶ s = 3.2ms=3.2m

The greatest height above the ground

\longrightarrow⟶ 85 + 3.2 m85+3.2m

\longrightarrow⟶ 88.2m88.2m

2)For the displacement motion-

\longrightarrow⟶ u = 0u=0

\longrightarrow⟶ g = 10m {s}^{ - 2}g=10ms

−2

\longrightarrow⟶ s = 88.2ms=88.2m

\longrightarrow⟶ {v}^{2} = {u}^{2} + 2gsv

2

=u

2

+2gs

\longrightarrow⟶ 0 + 2 \times 10 \times 88.20+2×10×88.2

\longrightarrow⟶ 17641764

3)Let t be time taken to reach the maximum height

For upward motion

\longrightarrow⟶ u = 8m {s}^{ - 1}u=8ms

−1

\longrightarrow⟶ v = 0v=0

\longrightarrow⟶ g = - 10m {s}^{ - 1}g=−10ms

−1

\longrightarrow⟶ t = \frac{v - u}{g}t=

g

v−u

\longrightarrow⟶ t = \frac{ - 8}{ - 10}t=

−10

−8

\longrightarrow⟶ t = 0.8st=0.8s