Question

A ball thrown vertically upward return to the thrower after 10s. Find
a) The velocity with which it was thrown up.
b) The maximum height it reaches.
c) Its position after 4s.
(Take g=10ms^2)


provide answer for this question if you don't know the answer so don't answer ​

Answer 1

Explanation:

(a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey. Hence, it has taken 3 s to attain the maximum height. Hence, the ball was thrown upwards with a velocity of 29.4 m s−1.

Answer 2

Answer:

$\bigstar{\bold{Initial\:velocity=50\:m/s}}$

$\bigstar{\bold{Maximum\:height=125\:m}}$

$\bigstar{\bold{Position\:after\:4\:s=120\:m}}$

Explanation:

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

• Time taken to return back = 10 s
• Final velocity = 0 m/s

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

• The velocity with which it was thrown upwards
• The maximum height it reaches
• Its position after 4 s

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

➺ By given the total time it takes to go up and come back down is 10 s.

➺ Hence time taken to reach the maximum height = 5 s

Initial velocity:

➺ Now we have to find the initial velocity of the body.

➺ By the first equation of motion,

   v = u + at

    where v = final velocity

                u = initial velocity

                a = acceleration

                t = time taken

➺ Substituting the data,

    0 = u - 10 × 5

➺ Here acceleration (a) = g = -10 m/s² since the motion of the body is in the direction opposite to that of acceleration due to gravity.

    0 = u + -50

    u = 50 m/s

➺ Hence the initial velocity of the object is 50 m/s.

    $\boxed{\bold{Initial\:velocity=50\:m/s}}$

➺ Now we have to find the maximum height attained by the body.

➺ By the third equation of motion,

    v² - u² = 2as

    where v = final velocity

                u  = initial velocity

                a = acceleration

                 s = displacement

➺ Substituting the data,

    0² - 50² = 2 × -10 × s

    -2500 = -20 s

     s = 2500/20

    s = 125 m

➺ Hence the maximum height attained by the body is 125 m.

    $\boxed{\bold{Maximum\:height=125\:m}}$

Position of the object after 4 s:

➺ Now we have to find the position of the body after 4 seconds.

➺ By the second equation of motion,

    s = ut + 1/2 at²

    where s = distance travelled

                u = initial velocity

                a = acceleration

                 t = time taken

➺ Substitute the data,

    s = 50 × 4 + 1/2 × -10 × 4²

    s = 200 - 80

   s = 120 m

➺  Hence position of the ball after 4 s is 120 m above the ground.

     $\boxed{\bold{Position\:after\:4\:s=120\:m}}$