John throws the ball straight upward and after 1 second, it reaches its maximum height then it does free fall motion which takes 2 seconds. Calculate the maximum height and velocity of the ball before it crashed the ground. (g=9.8 m/s^2)
Answers
Given:-
- Final Velocity of the ball = 0m/s
- Time = 1s.
- Acceleration Due to gravity = -9.8m/s²
To Find:-
- The maximum height reached by ball and Velocity of the ball before it crashed the ground.
Formulae used:-
- v = u + at
- S = ut + ½ × a × t²
Where,
- V = Final Velocity
- u = Initial Velocity
- S = Distance
- a = Acceleration
- t = Time.
Now,
First we find the initial Velocity with which the ball was thrown.
→ v = u + at
→ (0) = u + (-9.8) × 1
→ -u = -9.8
→ u = 9.8m/s
Hence The initial Velocity of the ball is 9.8m/s.
Now we will find the Distance travelled by the ba.
→ s = ut + ½ × a × t²
→ -s = 9.8 × 1 + ½ × (-9.8)× (1)²
→ s = 9.8 +4.9
→ s = 14.7m
Hence, The maximum Height reached by ball is 14.7m
Now,
During free fall, "u" become 0 So,
→ v = u + at
→ v = (0) + 9.8 × 2
→ v = 19.6m/s
Hence, The Velocity of the ball before it crashes is 19.6m/s.
Answer:
CASE 1:
AN OBJECT IS THROWN UPWARDS AND IT REACHES ITS MAXIMUM HEIGHT BY TAKING SOME TIME.
GIVEN:
FOR CASE 1 ,INITIAL VELOCITY = U M/S
FINAL VELOCITY = 0 M/ S
TIME =1 SECOND
ACCELERATION DUE TO g= 9.8 m/ s^2
TO FIND INITIAL VELOCITY WE USE THE FORMULA:
V=U-gt
=>0=U-98/10×1=>U= 98/10=9.8 M/ S.
NOW TO MAXIMUM HEIGHT IT REACHES WE USE THE FORMULA:
V^2=U^2-2gS
=>0^2-9.8^2=-2gS
=>-96.04=-2×98/10×S
=>96.04×10/98=S
=>9.8M=S
MAXIMUM HEIGHT IT REACHES IS 9.8 M.
CASE 2:
NOW THE BALL AFTER REACHED IT'S MAXIMUM HEIGHT IT THEN FREE FALL TO THE GROUND WITH POSITIVE ACCELERATION DUE TO GRAVITY.
GIVEN:
FOR CASE 2,
INITIAL VELOCITY = 0 M/S
TIME TAKEN 2 SECONDS.
FINAL VELOCITY = V M/S.
ACCELERATION DUE TO g = 9.8 M/S^2
HEIGHT = 9.8 M
WE HAVE TO FIND THE VELOCITY BEFORE IT HITS THE GROUND.
WE USE FORMULA :
V=U+gT=V=0+98/10×2=19.6 M/S
SO WE GOT FINAL VELOCITY = 19.6 M/S
HENCE VELOCITY BEFORE IT HITS THE GROUND IS
19.6 M/S.