Question

A ball is thrown vertically upwards with A a velocity 30m/s calculate the max height reached

Answer 1

Answer:

${\bold{\therefore h_{max}=45\:m}}$

Explanation:

${\bold{\huge{\underline{SOLUTION-}}}}$

• In the given question information given about a ball thrown vertically upward it means the ball is thrown at an angle of 90°.

• We have to find the maximum height.

$\underline \bold{Given :} \\ \\ \implies Initial \: velocity(u) = 30 \: m/s \\ \\ \implies Final \: velocity(v) = 0 \\ \\ \implies Angle \: of \: projection = 90 \degree \\ \\ \underline \bold{To \: Find :} \\ \\ \implies h_{max} = ?$

• According to given question :

$\: \: \: \: \: \boxed{ \bold{ First \: method : }}\\ \\ \bold{By \: formula \: of \: h_{max} : } \\ \implies h_{max} = \frac{ {u}^{2} { \sin^{2} \theta} }{2g} \\ \\ \implies h_{max} = \frac{ {30}^{2} \times { \sin^{2} 90\degree} }{2 \times 10} \\ \\ \implies h_{max} = \frac{900 \times {1}^{2} }{20} \\ \\ \bold{\implies h_{max} = 45 \: m}$

$\: \: {\boxed{ \bold{Alternate \: method : } }}\\ \\ \bold{By \:second\:equation\: of \: motion : } \\ \implies {v}^{2} = {u}^{2} + 2as \\ \\ \implies {0}^{2} = {30}^{2} + 2 \times( - 10 )\times s \\ \\ \implies - 900 = - 20 \times s \\ \\ \implies s = \frac{ - 900}{ - 20} \\ \\ \bold{\implies s = 45 \: m}$

Answer 2

Solution:

Given:

➜ A ball is thrown vertically upwards with A a velocity 30m/s.

Find:

➜ Calculate the max height reached. ( Hₘₐₓ = ? )

According to given question:

➜ The answer to this question can be done in two method.

1. Method (1) using Hₘₐₓ formula.
2. Method (2) using 2nd equation of motion.

Method-(1)

Using Hₘₐₓ formula:

➜ Hₘₐₓ = u² sin² θ / 2g

➜ Hₘₐₓ = 30² × sin² 90° / 2 × 10

➜ Hₘₐₓ = 900 × 12 / 100

➜ Hₘₐₓ = 45 m

Method-(2)

Using second equation of motion:

➜ v² = u² + 2 as

➜ 0² = 30² + 2 × (-10) × 5

➜ -900 = -20 × 5

➜ s = -900 / -20

➜ s = 45 m

Know terms:

➜ (θ) => theta.

➜ (°) => degree.

➜ (v) => initial velocity.

➜ (u) => final velocity.

#AnswerWithQuality

#BAL