Question

A ball is projected at an angle of 90° from horizontal with a inital velocity 120 m/s . Find the maximum height ball reached and the time taken to reach maximum height.​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Height=720\:m}}}$

$\green{\tt{\therefore{Time\:taken=12\:sec}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Inital \: velocity(u) = 120 \: m/s \\ \\ \tt: \implies Angle \: of \: projection = 90 \degree \\ \\ \tt: \implies Acceleration(a) = 10 \: m/{s}^{2} \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Maximum \: height = ? \\ \\ \tt: \implies Time \: taken(t) =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {0}^{2} = {(120)}^{2} + 2 \times ( - 10) \times s \\ \\ \tt: \implies 0 = 14400 - 20 \times s \\ \\ \tt: \implies - 20 \: s = - 14400 \\ \\ \tt: \implies s = \frac{ - 14400}{ - 20} \\ \\ \green{\tt: \implies s = 720 \: m} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies 0 = 120 + ( -10) \times t \\ \\ \tt: \implies - 10t = - 120 \\ \\ \tt: \implies t = \frac{ - 120}{ - 10} \\ \\ \green{\tt: \implies t = 12 \: sec}$

Answer 2

$</p><p>\huge{\fbox{\fbox{\mathfrak {\rightarrow{\leftarrow {\green{ANSWER \: - }}}}}}}$

The maximum height reached by the ball is 720 metres and the time taken is 12 seconds.

$</p><p>\orange{\underline {\underline {\tt{Step-By-Step-Explaination \: : }}}}$

$</p><p>\huge{\boxed{\boxed{\underline{\bold{\blue{\mathfrak{QUESTION \: : }}}}}}}$

A Ball Is Projected At An Angle Of 90° With The Horizontal With A Initial Velocity 120 m/s.

To Find

• Maximum Height Reached By The Ball

• Time Taken To Reach The Maximum Height

PART A

$</p><p>\red{\underline{\underline{\bigstar{\tt{Formula \: Used \: - }}}}}$

$</p><p>\huge{\fbox{\fbox{\rightarrow{\tt{\blue{V^2 \: = U^2 \: + 2A×S }}}}}}$

Here, in this question, the ball finally reaches the ground hence:

• V = 0 m/s

• U = 120 m/s

• g = -10 m/s approx

$</p><p>\red{\underline{\underline{\bigstar{\tt{To \: Find \: - \: S }}}}}$

$</p><p>\green{\bold{\tt{\boxed{\boxed{\therefore{0^2 \: = 120^2 - 10S }}}}}}$

$</p><p>\huge{\blue{\fbox{\fbox{\bigstar{\mathfrak{S \: = 720 \: METRES.}}}}}}$

PART B

$</p><p>\red{\underline{\underline{\bigstar{\tt{Formula \: Used \: - }}}}}$

$</p><p>\huge{\fbox{\fbox{\rightarrow{\tt{\orange{V \: = U \: + AT }}}}}}$

In this Case, the same values are used again...

Here, in this question, the ball finally reaches the ground hence:

V = 0 m/s

U = 120 m/s.

g = -10 m/s approx

$</p><p>\green{\bold{\tt{\boxed{\boxed{\therefore{0 \: = 120-10T }}}}}}$

$</p><p>\huge{\blue{\fbox{\fbox{\bigstar{\mathfrak{T \: = 12 \: SECONDS.}}}}}}$

$</p><p>\huge{\red{\mathfrak {\bold{\rightarrow{\leftarrow{HOPE \: THIS \: HELPS!!! }}}}}}$