A ball thrown vertically upwards with a speed of 19.6 ms^-1 from the top of a tower returns to the earth in 6 seconds. Find the height of the tower. (g = 9.8 ms^-2 ).
Answers
Given that,
Initial velocity u=19.6m/s
Final velocity v=0
Time t=6sec
The acceleration is
We know that,
v=u+at
0=19.6+a×6
a=−3.26m/s
2
Now, the height is
From equation of motion
s=ut+
2
1
at
2
s=19.6×6−
2
1
×3.26×36
s=58.9
s=59m
Hence, the height of the tower is 59 m
A ball thrown vertically upwards with a speed of 19.6 m/s from the top of a tower returns to the earth in 6 seconds.
We have to find the height of the tower.
From above data we have; initial velocity of the ball is 19.6 m/s, final velocity of the ball is 0 m/s (as highest point final velocity becomes 0) and time is 6 sec.
Using the First Equation Of Motion i.e. v = u + at
Substitute the known values in the above formula,
→ 0 = 19.6 + a(6)
→ -19.6 = 6a
→ -3.27 = a
(Negative sign shows retardation)
Therefore, the acceleration of the ball is 3.27 m/s².
Now,
Using the Third Equation Of Motion i.e. s = ut + 1/2 at²
Substitute the known values,
→ s = 19.6(6) + 1/2 × (-3.27)(6)²
→ s = 117.6 - 18(3.27)
→ s = 117.6 - 58.86
→ s = 58.74
→ s = 59 (approx.)
Therefore, the distance covered by the ball is 59 m.
OR
Using First Equation Of Motion, v = u + at
→ 0 = 19.6 + (-9.8)t
→ t = 2
Using the Second Equation Of Motion, s = ut + 1/2 at²
a = g = 9.8 m/s² (given in question)
Substitute the known values in the above formula,
u = 0 and t = 6-2 = 4
→ s = 0(2) + 1/2 (9.8)(4)²
→ s = 78.4 m
Height of tower = 78.4 - 19.6 = 58.8 = 59 m
OR
s = 19.6(2) + 1/2 × 9.8(4)²
s = 58.8 = 59 m