Question

A ball was thrown vertically upward with an initial velocity of 16 m/s and it returned to the ground 6 s later. How high did the ball rise?

Answer 1

Explanation:

A ball's thrown upwards from the top of a building with an initial velocity of 19.6 m/s. The ball reaches the ground after 5 seconds. What will be the height of building?

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7 Answers

Satya Parkash Sud, M.Sc. Physics & Nuclear Physics, University of Delhi (1962)

Answered 3 years ago · Author has 4K answers and 10.1M answer views

Let the height of the building is H metre. The ball is thrown upwards with an initial velocity u = 19.6 m/s. Due to being thrown up the ball will travel upwards till its upwards velocity becomes zero. The additional height h gained by the ball can be calculated using the relation;

v² - u² = 2 g h.

In our case u = + 19.6 m/s ( we are taking upwards direction as positive), v= 0 m/s (the ball will keep travelling upwards till its velocity becomes zero); g = - 9.8 m/s², h= height at which velocity v becomes zero. Substituting the values in the equation,we get,

0² - 19.6² = - 2×9.8×h, is h= 19.6 m. So the ball now falls from a height = H + 19.6 m.

Let us find the time the ball takes to reach 19.6 m above the top of the tower. Using the relation,

v = u + a t; ===> 0 = 19.6 - 9.8×t, or t = 2 s .

Now the total time the ball takes to reach the earth is 5s. Of this the ball takes 2s to reach the 19.6m above the top of the tower. So the time taken to fall to the ground from the highest point is 3 s. When the ball is at the highest point and starts its descent to the earth its initial velocity is u = 0 m/s;. g = +9.8 m/s²; t = 3s. Using the relation,

s = u t + ½ g t² , we get on substituting the values

s = 0×3 + ½ ×9.8 ×3²= 44.1 m

Therefore the height of the tower = 44.1 m - 19.6 m= 24.5m.

We could have obtained the same answer using the relation;

s = u t + ½ g t², where u= 19.6 m/s; t = 5s; g = -9.8 m/s² ( we are taking g as - 9.8 m/s² as g is downwards while u is upwards and taken as positive),so

s = 19.6×5 + ½ (-9.8)×5²= 98 m - 4.9×25= 98m-122.5 = -24.5 m. The sign of s being negative only means that it is the position of the ball after 5 s of its being thrown up with top of the building being as origin and upwards direction as positive.

So -24.5 m merely means the depth of the base of the tower ie the ground from the top of the tower. So wrt the ground height of the tower = 24.5 m.

Answer 2

Answer:

$\sf{s=92.1 m \approx 92 m}$

Explanation:

Question:

A ball was thrown vertically upward with an initial velocity of 16 m/s and it returned to the ground 6 s later. How high did the ball rise?

Given:

• Initial velocity = 16m/s
• Time taken for the ball to reach the ground = 6s
• Acceleration = $\sf{-9.8m/s^2}$

To find:

• Height or the distance of the ball

Note:

→Total time taken for the ball to move upwards and downwards = 6s

→So, the time taken to move upwards:

$\sf\dfrac{6}{2}$

★Time taken to reach upwards = 3s

Formula used:

$\sf{s=ut+1/2at^2}$

Where,

• s = Distance
• u = Initial velocity
• a = Acceleration
• t = Time taken

Solution:

◎Apply the formula:

●$\sf{s=ut+1/2at^2}$

●$\sf{s=16(3)+1/2(9.8)3^2}$

●$\sf{s=48+4.9(9)}$

●$\sf{s=48+4.9(9)}$

●$\sf{s=48+44.1}$

●$\sf{s=48+44.1}$

●$\sf{s=92.1 m \approx 92 m}$