A body is thrown vertically upwards from ground with a velocity u. It passes a point at a height h above the ground at time t1 while going up and at a time t2 while falling down. (t1 and t2 are measured from the instant of projecting body upwards).Then the relation between u, t1 and t2 is :
Answers
A body is thrown vertically upward from ground with a velocity u. time taken by body to reach height h is t1.while going up and at a time t2 while falling down
Let time taken to reach maximum height H by body is T.
at maximum height velocity of body is 0
from formula, v = u + at
v= 0, t = T and a = -g
so, T = u/g......(1)
time of flight = time taken to reach h + time taken to reach maximum height and then to reach ground
or, Time of ascending + time of descending = t1 + t2
or, T + T = t1 + t2 [ motion under gravity, time of ascending= time of descending]
or, 2T = t1 + t2
from equation (1),
t1 + t2 = 2u/g
Answer:
Explanation:
When a body, that is thrown vertically with velocity u, reaches a vertical distance h from ground at time t1,
Thus, we will have, h = u×t1 - (1/2)g×t1² --- (1)
If H is maximum the height it reaches with time T, then we will have
H = u2 / (2g) --- (2)
T = u/g ............................... (3)
During descending, if the object crosses the same spot which is at a height h above the ground, then the vertical distance travelled during ascending will be ( H-h ). Thus -
(H - h) = (1/2) gt² ---(4)
where t is the time taken to travel from highest point to the distance (H-h) .
If t2 is the time for the object, when it starts from the ground, and reaches maximum height and again crosses the spot which is at a distance h from ground, then we have,
t = (t2 - T ) --- (5)
Writing Eqn.(4) by substituting for H, h and t using respective eqns. (2), (1), (3) and (5)
[ u2 / (2g) ] - u×t1 + (1/2)g×t12 = (1/2)g [ t² - (u/g) ]² --- (6)
After simplifying equation (6), we will get,
t1 + t2 = 2u/g