Date:
5. A ball is dropped off a very tall canyon ledge. Gravity accelerates the bar
at 9.8 mis. How fast is the ball traveling after 5 seconds?
Answers
Given :
- Initial velocity of ball, u = 0 [since, ball is dropped from a very high canyon ledge]
- Acceleration due to gravity, a = 9.8 m/s²
To find :
- velocity of ball after time (t) of 5 second of being dropped, v = ?
Formula required :
- First equation of motion
v = u + a t
[ Where v is final velocity, u is initial velocity, a is acceleration, t is time taken ]
Solution :
Using the first equation of motion
→ v = u + a t
→ v = ( 0 ) + ( 9.8 ) ( 5 )
→ v = 9.8 × 5
→ v = 49 m/s
Therefore,
- Final velocity of ball after 5 seconds will be 49 m/s.
Three equations of motion :
- First equation of motion
v = u + a t
- Second equation of motion
s = u t + 1/2 a t²
- Third equation of motion
2 a s = v² - u²
[ Where v is final velocity, u is initial velocity, a is acceleration, t is time taken, s is distance covered ]
Answer:
✡ Correct Question ✡
➡ A ball is dropped off a very tall canyon ledge. Gravity accelerates the barat 9.8 m/s². How fast is the ball traveling after 5 seconds?
✡ Given ✡
➡ A ball is dropped off a very tall canyon ledge. Gravity accelerates the barat 9.8 m/s².
✡ To Find ✡
➡ The ball traveling after 5 second ?
✡ Formula Used ✡
➡ Newton's 1st Equation Motion :-
✡ Solution ✡
➡ By using 1st equation of motion:-
✴ v = u + at ✴
According to the question:-
=> v = (0) + (9.8) + (5)
=> v = 9.8 × 5
=> v = 49 m/s
∴ Final velocity of ball after 5 second is 49 m/s.
_____________________
✡ Additional Information ✡
▶ Newton's 1st law of motion:-
✏ The first law defines the force qualitatively.
✴ v = u + at ✴
▶ Newton's 2nd law of motion:-
✏ The second law offers a quantitative measure of the force.
✴ s = ut + at²
▶ Newton's 3rd law of motion:-
✏ The third asserts that a single isolated force doesn't exist.
✴ 2as = v² - u² ✴
Explanation: