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If a ball is thrown upward with speed of 20 m/s
& constant downward acceleration is 10 m/s2
then calculate total distance travelled (in
metres) when it reaches ground.
Answers
Given :
Initial velocity of ball = 20m/s
Acc. due to gravity = 10m/s²
To Find :
Total distance covered by ball.
Solution :
★ For a body thrown vertically upward, g is taken negative.
Since acceleration due to gravity has constant magnitude throughout the motion, we can apply equation of kinematics to solve this question.
First of all we need to find maximum upward distance covered by the ball
Applying third equation of motion;
➠ v² - u² = 2gH
- v final velocity
- u initial velocity
- g denotes acceleration
- H denotes height
At maximum point, v = 0
➠ 0² - 20² = 2(-10)H
➠ H = -400/-20
➠ H = 20m
Maximum height attained by the ball is 20m.
Total distance covered by ball :-
⭆ d = upward + downward
⭆ d = 20 + 20
⭆ d = 40 m
Given :
Initial velocity of ball = 20m/s
Acc. due to gravity = 10m/s²
To Find :
Total distance covered by ball.
Solution :
★ For a body thrown vertically upward, g is taken negative.
Since acceleration due to gravity has constant magnitude throughout the motion, we can apply equation of kinematics to solve this question.
First of all we need to find maximum upward distance covered by the ball..
Applying third equation of motion;
➠ v² - u² = 2gH
v final velocity
u initial velocity
g denotes acceleration
H denotes height
At maximum point, v = 0
➠ 0² - 20² = 2(-10)H
➠ H = -400/-20
➠ H = 20m
Maximum height attained by the ball is 20m.
Total distance covered by ball :-
⭆ d = upward + downward
⭆ d = 20 + 20
⭆ d = 40 m