Question

. A ball is thrown upward at 25 m/s from the ground.
• What is the initial velocity of the ball?
• What is the acceleration of the ball?
• What is the ball’s velocity after 2 seconds?
• What is the ball’s velocity after 4 seconds?
• What is the maximum height of the ball?
• How long until the ball hits the ground?
• When is the magnitude of the velocity 5 m/s?
• What distance has the ball travelled after 5 seconds?
• What is the average velocity and average speed of the ball after 5 seconds?
• Another ball is thrown one second later. What speed does it need to hit the ground simultaneously with the first ball?

Answer 1

Initial Velocity-25 m/s

Acceleration=-g m/s²=-9.81m/s²

Velocity of ball after 2 seconds-

v=u+at

v=25-(9.8)2=25-19.6=5.4 m/sec

Distance ball had travelled after 5 sec

It is the distance travelled by the ball when its reached maximum heigh(v=u+at)

(0=25-g(t))

(-25=-gt) From here we get the time as 25/g

which is approximately 2.5 sec

we know time taken by the ball to reach is maximum height is equal to the time taken by the ball to come down. ......(1)

So after 5 sec the ball will return to its initial position

$(s1)=ut+1/2at²=25(2.5)+(1/2)(-g)(2.5)². (calculate yourself)

This distance needs to be doubled. [(1)]

Average velocity=0 as displacement =0

Average speed=Total distance/Total time

=2$(2S1)/5(total time). [calculate yourself]

And the last part of the question

see we know the time of flight of the ball(5sec)

know we know that the first ball reaches the maximum distance are 2.5 seconds

So here is the catch We have to solve this part by assuming a velocity of second ball then as the two balls should travel equal distance ball 1 in 5 sec and ball 2 in 4 sec

now put in the equation

s(Total distance)=u2(4)+1/2(-g)t²

Total distance we can find its 2$(2s1)

and then pur it in the above equation

I hope calculation part u can do it urself

If u have any problem related to it kindly let me know in the comments section

;-)

Answer 2

Given:

A ball is thrown upward at 25 m/s from the ground.

To Find:

The answers to the given questions.

Answers:

1). Initial velocity is 0 m/s.

2). Acceleration of the ball is 9.8 m/s.(acceleration due to gravity.)

3). The velocity of the ball after 2 seconds is 19.6 m/s.(velocity = acceleration × time).

4). The velocity of the ball after 4 seconds is 39.2 m/s.(velocity = acceleration × time).

5). The maximum height of the ball is 156.8m. (distance = velocity × time).

6). The ball hits the ground in 5 seconds.

7). At 0.5 seconds the magnitude of the velocity is 5 m/s. (time = acceleration × velocity).

8). The distance has the ball traveled after 5 seconds is 160m.

9). Average velocity and speed is 30 m/s.

10). 26m/s.