Question

49 ÷ 9.8 = 5
Write in steps​

Answer 1

Answer:

5

Step-by-step explanation:

$\frac{49}{9.8} = \frac{49 \times 10}{98} = 5$

Answer 2

A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate:

a. the max height to which it rises.

b. total time it takes to return to surface of the earth.

Answer:

\bigstar{\bold{Maximum\:height=122.5\:m}}★Maximumheight=122.5m

\bigstar{\bold{Total\:time=10\:s}}★Totaltime=10s

Explanation:

\Large{\underline{\underline{\bf{Given:}}}}

Given:

Initial velocity of ball (u) = 49 m/s

Final velocity of ball (v) = 0 m/s

\Large{\underline{\underline{\bf{To\:Find:}}}}

ToFind:

Maximum height it attains = Distance travelled (s)

Time taken (t)

\Large{\underline{\underline{\bf{Solution:}}}}

Solution:

a. Maximum height it attains:

→ By the third equation of motion we know that,

v² - u² = 2as

→ Here acceleration on the body is acceleration due to gravity which acts in the direction opposite to the motion of the body.

→ Hence a = -g = -9.8 m/s²

→ Substitute the given datas in the above equation,

0² - 49² = 2 × - 9.8 × s

s = -2401/-19.6

s = 122.5 m

→ Hence the maximum height attained by the body = 122.5 m

\boxed{\bold{Maximum\:height=122.5\:m}}

Maximumheight=122.5m

b.Total time taken to return back

→ By the first equation of motion, we know that

v = u + at

where a = -g = -9.8 m/s²

→ Substituting the datas given, we get the value of time.

0 = 49 + -9.8 × t

-9.8 × t = -49

t = 49/9.8

t = 5 s

→ Since the object has to travel up and down, the total time taken would be 5 + 5 = 10 s

→ Hence the total time taken for the ball to return back = 10 s

\boxed{\bold{Total\:time=10\:s}}

Totaltime=10s

\Large{\underline{\underline{\bf{Notes:}}}}

Notes:

→ The three equations of motion are:

v = u + at

s = ut + 1/2 × a × t²

v² - u² = 2as