Q. A ball is thrown vertically upward from the top of a tower with a velocity of 10 m/sec. If the ball falls on the ground after 5 seconds the height of the tower will be.
Q. In the above question, what maximum height above the tower will the ball attain?
Q. In the above question, the total distance traveled by the ball before it returns to the ground is
Q. In the above question, at what time will it reach the maximum height?
Q. In the above question, with what velocity will the ball strike the ground?
Answers
Answer:
s=+ut+
2
1
at
2
s=+(10)(5)+
2
1
(10)(5)
2
s=+50+
2
1
×10×25
s=50+50m=100m
Note:- v = u + at
Q1. We know that,
s = ut+ 1/2 at ²
s = (10)(5)+ 1/2(10)(5) ²
s = 50+ 1/2 ×10×25
s = 50+50m=100m
Q2. Let A ball is thrown upward direction with speed and the ball reaches h height from the top of tower, where velocity is v And after reaching highest point, ball falls downward. At horizontal line of top of tower, ball again gains velocity v as shown in figure. Now ball is moving downward and falling h height. Let the velocity of ball h height below from the top of tower is v'.
Case 1:
In case of upward motion. use formula, v² = u² + 2aS Here, a = -g, S = h
Then, v² u² - 2gh v= √(u² - 2gh) •••••••(1)
Case 2:
In case of downward motion, Use same formula, v² = u² + 2aS v¹² = u² + 2(-g)(-h) = u² + 2gh v' = √(u² + 2gh). •••••••(2)
Here, a = -g, S = -h
Now, A/C to question, velocity of ball h height below = 2 x velocity of ball h height below
√(u² + 2gh) = 2√(u² - 2gh) Squaring both sides,
u² + 2gh = 4(u² - 2gh)
=> u² + 2gh = 4u² - 8gh -3u² = - 10gh
=> u² = 10gh/3 •••••(1)
ball is throwing upawrd, at maximum height velocity of ball will be zero. Let ball reaches Maximum height H.
v² = u² + 2aS Here, v = 0, u² = 10gh/3, a = -g and S = H Now, 0 = 10gh/3 - 2gH
=> H = 5h/3
Hence, answer is maximum height, H = 5h/3