A ball is thrown vertically upwards with a velocity of 25 ms from the top of a tower ol height 30 m. How long will it travel before it hits ground 6s (2) 5
Answers
Answer:
t = 3 sec
Explanation:
Given ;
Initial velocity ( u ) = 25 m / sec
Height ( h ) = 30 m
Final velocity will be 0 m / sec
we have to find time ( t )
From second equation of motion we have
h = u t - 1 / 2 g t² [ Acceleration against gravity ]
putting values
[ Here displacement will be in negative ]
- 30 = 25 t - 1 / 2 × 10 × t² [ Taken g = 10 ]
- 30 = 25 t - 5 t²
- 6 = 5 t - t²
t² - 5 t - 6 = 0
( t - 3 ) ( t + 2 ) = 0
t = 3 sec or t = - 2 sec .
Time can't in negative ;
Thus we get time t = 3 sec .
Answer:
Given :-
Initial velocity (u) = 25 m/s
Height (h) = 30 m
We know that it is thrown from the top of tower. So,
Final velocity (v) = 0 m/s
Acceleration (g) = 10 m/s²
Find :-
Time = ?
Solution :-
Using second equation of motion
S = ut - 1/2 gt²
-30 = 25t - 1/2 × 10 × t²
5 = 25t - t²
t² - 5t -30 = 0
By Factorising Method
(t - 3) (t + 2)