Question

a boy is standing on the edge of the roof
at height 78.4 m. He throws a ball
vertically upwards with velocity 19.6ms
After how much time the ball will
reach the ground?
Take a = 9.8m/s^2 downward


pls solve this que it's urgent...​

Answer 1

Answer:

The ball will reach the ground after $6.47$ sec.

Explanation:

Concept:

The total time $=$ the length of time it took the ball to reach the roof $+$ the amount of time it takes for a ball to velociously fall from a roof to the ground $-19.6 m/s$.

Given:

A boy is standing at height $78.4 m$.

He launches a ball quickly and vertically upward. $19.6ms$

To find:

We have to find that after how much time the ball will reach the ground.

Solution:

Time to reach the roof,

$=\frac{ 2(19.6)}{9.8} \\=4s$

The time it took to get to the ground is now discovered,

initial velocity$=19.6$

Height$=78.4 m$

Final velocity and time is,

$v^2 = u^2 + 2gh$

Substitute the values:

$v=\sqrt{(19.6^2 + 2(78.4)(9.8)}$

$v = 43.8$

Now apply formula to find time,

$v = u + at\\43.8 - 19.6 = 9.8t\\ t=2.47s$

Therefore the total time,

$4+2.47= 6.47s$

Hence, the ball will reach the ground after $6.47$ sec.

#SPJ3

Answer 2

Answer:

The time taken by the ball to reach the ground is 6.47s.

Explanation:

Given:

Height(s) = 78.4m

Initial velocity(u) = 19.6m/s

Acceleration(a) = -9.8m/s²

To find:

Time taken to reach the ground(t) = ?

Step 1: To find the time taken by the ball to reach the maximum height.

We use the first equation of motion given as follows:

v = u + at₁

0 = 19.6 + (-9.8)t₁

0 = 19.6 - 9.8t₁

9.8t₁ = 19.6

t₁ = 2s

Step 2: To find the maximum height reached by the ball when thrown upwards.

We use the third equation of motion given as v² = u² + 2as

Substituting the values, we get

0 = (19.6)² +2(-9.8)s

0 = (19.6)² - 19.6s

19.6s = 19.6×19.6

s = 19.6m

Step 3: To find the total time taken to reach the ground.

The ball is thrown from a roof which is at a height of 78.4m. Therefore,

total height = 78.4 + 19.6

total height = 98m

On reaching the maximum height, the velocity will become zero, i.e., u=0.

Using the second equation of motion, we get

s = ut₂ + (1/2)at₂²

98 = 0 + (1/2)(9.8)t₂²

98 = 4.9t₂²

t₂² = 20

t₂ = √20

t₂ = 4.47

Therefore, the total time taken is

t = t₁ + t₂

t = 2 + 4.47

t = 6.47s

Hence, the ball will reach the ground after 6.47s.

#SPJ1