A ball is thrown from a tower of height 40m with a velocity of 40m/s and another ball is thrown with a velocity of 24m/s from the bottom of the tower. When and where will the 2 bodies meet with respect to the first ball?
Answers
Answer:
The balls will meet at a height of 77.5 m after a time of 5 seconds.
Explanation:
Given:
A ball is allowed to fall from the top of a tower of height 200 m
Another ball is thrown vertically upwards with a height of 40 m/s
To Find:
Height at which the balls will meet
Time at which the balls will meet
Solution:
Let us assume that the balls meet a time t seconds and a distance of s m from the ground.
Therefore at the time of meeting,
Distance travelled by ball which was dropped down = (200 - s) m
Distance travelled by the ball thrown upward = s m
For the ball moving downwards,
By the second equation of motion we know that,
s = ut + 1/2 × a × t²
where s = distance travelled
u = initial velocity
a = g = acceleration due to gravity
t = time taken
Substitute the data,
(200 - s) = 0 × t + 1/2 × 9.8 × t²
200 - s = 4.9 t²
s = 200 - 4.9 t²-----(1)
For the ball thrown upwards,
s = ut + 1/2 × at²
Here
a = -g ( since the direction of motion of the object is opposite to that of acceleration due to gravity.)
Substitute the data,
s = 40 × t + 1/2 × -9.8 × t²
s = 40t - 4.9t²-------(2)
Since LHS of equation 1 and 2 are equal, RHS must also be equal.
Hence,
200 - 4.9t²= 40t - 4.9t²
200 = 40 t
t = 200/40
t = 5 s
Hence the balls will meet after a time of 5 seconds.
Substitute the value of t in equation 1,
s = 200 - 4.9 × 5²
s = 200 - 122.5
s = 77.5 m
Hence the balls will meet at a height of 77.5 m.
Answer:
velocity after 2 sec.
v=40+(−10)t
=40−10×2= 20m/s
Explanation:
v = u + a×t