A ball is thrown vertically upwards with a speed of 20m/s from a tower of height 40 m. Find the maximum height from the ground covered by the ball and total time taken to reach the ground after release from the tower.
Answers
Answer:
Maximum height of the ball from ground = 40 meter
total time taken to reach ground
= 2(1 + √2) seconds
Step by step explanations :
Given that,
A ball is thrown vertically upwards with a speed of 20m/s from a tower of height 40 m
here,
initial velocity of the ball = 20 m/s
height of the tower = 40 m
let the height of its maximum height reached by ball from tower be h and
from ground be H
so,
H = h + 40
now,
at the maximum height velocity will 0
so,
now we have ,
Initial velocity(u) = 20 m/s
final velocity(v) = 0 m/s
height = h
gravitational acceleration(g) = -10 m/s²
[upward]
so,
by the gravitational equation of motion
v² = u² + 2gh
putting the values,
0² = (20)² + 2(-10)h
-20h = -400
h = -400/-20
h = 20 m
and we have,
H = h + 20
where,
H is maximum height of the ball from ground
so,
H = 20 + 20
H = 40 m
so,
Maximum height of the ball from ground = 40 meter
_____________________
Now,
to find time taken,
we know that,
time taken by the ball to reach its height after throwing it
= u/g
where,
u is the initial velocity
so,
time taken by ball to reach ground
= time taken (to reach its maximum height + time taken to reach ground from its maximum height)
so,
time taken to reach its maximum height
= 20/10
= 2 seconds
and we know that,
speed of ball at it's maximum height= 0
so,
u = 0 m/s
g = 10 m/s [downward]
H = 40 m
by the gravitational equation of motion
H = ut + ½ gt²
putting the values
40 = 0(t) + ½ × 10t²
5t² = 40
t² = 40/5
t² = 8
t = 2√2 s
so,
total time taken to reach ground
= 2s + 2√2s
= 2(1 + √2) seconds
_____________________
Maximum height of the ball from ground = 40 meter
total time taken to reach ground
= 2(1 + √2) seconds
Answer:
Maximum height = 40 m
total time taken = 2(1 + √2) s
Step by step explanations :
Given,
Initial velocity(u) = 20 m/s
final velocity(v) = 0 m/s
height = h
g = -10 m/s²
v² = u² + 2gh
0 = 20² + 2(-10)h
-20h = -400
h = -400/-20
h = 20 m
and we have,
H = h + 20
where,
H is maximum height of the ball from ground
so,
H = 20 + 20
H = 40 m
so,
Maximum height of the ball from ground = 40 meter
now,
time taken by the ball to reach its maximum height from tower
= u/g
= 20/10
= 2 s
and we know that,
speed of ball at it's maximum height
= 0 m/s
so,
u = 0 m/s
g = 10 m/s
H = 40 m
H = ut + ½ gt²
putting the values
40 = 0(t) + ½ × 10t²
5t² = 40
t² = 40/5
t² = 8
t = 2√2 s
so,
total time taken to reach ground
= 2 + 2√2
= 2(1 + √2) seconds