Question

6.
A particle is fired vertically upward with a speed
of 9.8 km s. Find the maximum height attained
by the particle. Radius of earth = 6400 km and g
at the surface = 9.8 ms 2. Consider only earth's
gravitation.

Answer 1

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

${\bold{\therefore Distance=4900\:km}}$

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

• In the given question information given about a particle is fired vertically upwards with a speed of 9.8 km/s.

• We have to find the maximum height attained.

$\underline \bold{Given : } \\ \implies Initial \: velocity(u) = 9.8 \: km / s \\ \\ \implies Acceleration(a) = - 9.8 \: m {s}^{2} \\ \\ \implies Radius \: of \: earth = 6400 \: km \\ \\ \underline \bold{To \: Find: } \\ \implies Distance(s) = ?$

• According to given question :

$\bold{u = 9.8 \times 1000 = 9800 \: m/s} \\ \\ \bold{Using \: second \: equation \: of \: motion : } \\ \implies {v}^{2} = {u}^{2} + 2as \\ \\ \bold{Putting \: given \: values : } \\ \implies {0}^{2} = {9800}^{2} + 2 \times ( - 9.8) \times s \\ \\ \implies - ({9800})^{2} = - 2 \times 9.8 \times s \\ \\ \implies s = \frac{ \cancel{9800} \times 9800}{ \cancel2 \times \cancel{9.8}} \\ \\ \bold{\implies s = 4900000 \: m \: or \: 4900 \: km}$

Answer 2

Answer:

○ Distance= 4900 km

Explanation:

Given :

□ Initial velocity=9.8 km/s=9800 m/s

□ Acceleration = 9.8 m/s^{2}.

■ Using second equation of motion :

$\to {v}^{2} = {u}^{2} + 2as \\ \\ \to 0 = {9800}^{2} - 2 \times 10 \times s \\ \\ \to s = \frac{9800 \times 9800}{2 \times 9.8} \\ \\ \bold{\to s = 4900000 \:m} \\ \\ \bold{\therefore distance \: travelled = 4900 \: km}$