A ball is thrown vertically upwards from the top of the tower with an initial velocity of 19.6ms-1 . The ball reaches the ground after 5s. Calculate:(i) the height of the tower (ii) the velocity of ball on reaching the ground . Take g=9.8ms-2
Answers
Answer:
s=ut-1/2gt*2.
Explanation:
s=19.6*5-1/2*9.8*(5)*2. 0=19.6-1/2*9.8*5. 0=19.6-24.5. 0=5. h=5. 2.v=u-gt. 0=19.6-9.8*5. v=28.4m/s. hope it will help you
height of the tower = 24.5 m & the velocity of ball on reaching the ground . = 29.4 m/s &
Explanation:
initial Velocity = 19.6 m/s upward
Velocity at top = 0
a = -g = -9.8 m/s²
0 = 19.6 + (-9.8)t
=> t = 2 sec
Distance Covered going upward using V² - U² = 2aS
S = (0² - 19.6²)/(2 * (-9.8)) = 19.6 m
The ball reaches the ground after 5s
=> time taken from top to ground = 5 - 2 = 3 m
Now taking velocities downward
u = 0
V = Velocity at ground
a = g = 9.8 m/s²
V = 0 + 9.8 * 3
=> V = 29.4 m/s
the velocity of ball on reaching the ground . = 29.4 m/s
Distance Covered from top to ground = ( 29.4² - 0²)/(2 * 9.8)
= 44.1 m
Height of Tower = 44.1 - 19.6 = 24.5 m
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