Question

A ball is thrown vertically upward rises to a height of 45 m. Calculate

(a) Initial velocity of ball

(b) Total time of flight

(c) Position of ball at time t = 5 secon​

Answer 1

Initial velocity of the ball (u) = u m/s

Total height (s) = 45 m

As the ball thrown against the gravity , so

Acceleration (a) = -10 m/s²

As the ball attains rest at extreme position  , so

Final velocity (v) = 0 m/s

(a) Initial velocity of ball

Apply 3rd equation of motion ,

$:\implies \sf v^2-u^2=2as$

$:\implies \sf (0)^2-u^2=2(-10)(45)$

$:\implies \sf 0-u^2=-900$

$:\implies \sf u^2=(30)^2$

$:\implies \boxed{\sf u=30\ m/s}$

(b) Total time of flight

Apply 1st equation of motion ,

$:\implies \sf v=u+at$

$:\implies \sf 0=30+(-10)t$

$:\implies \sf 10t=30$

$:\implies \boxed{\sf t=3\ s}$

For coming back to initial position where it is thrown , it again takes 3 s , so

Total time of flight = 3 + 3 = 6 s

(c) Position of ball at time t = 5 seconds

Ball takes 3 s to reach max. height and again falls back from there ( from rest )

So we need to find position of ball at 2 s (5-3)

$:\implies \sf s=ut+\dfrac{1}{2}at^2$

$:\implies \sf s=(0)(2)+\dfrac{1}{2}(10)(2)^2$

$:\implies \boxed{\sf s=20\ m}$

Distance from the ground = 45 m - 20 m = 25 m

Answer 2

Answer:

Initial velocity of the ball (u) = u m/s

Total height (s) = 45 m

As the ball thrown against the gravity , so

Acceleration (a) = -10 m/s²

As the ball attains rest at extreme position , so

Final velocity (v) = 0 m/s

(a) Initial velocity of ball

Apply 3rd equation of motion ,

:\implies \sf v^2-u^2=2as:⟹v

2

−u

2

=2as

:\implies \sf (0)^2-u^2=2(-10)(45):⟹(0)

2

−u

2

=2(−10)(45)

:\implies \sf 0-u^2=-900:⟹0−u

2

=−900

:\implies \sf u^2=(30)^2:⟹u

2

=(30)

2

:\implies \boxed{\sf u=30\ m/s}:⟹

u=30 m/s

(b) Total time of flight

Apply 1st equation of motion ,

:\implies \sf v=u+at:⟹v=u+at

:\implies \sf 0=30+(-10)t:⟹0=30+(−10)t

:\implies \sf 10t=30:⟹10t=30

:\implies \boxed{\sf t=3\ s}:⟹

t=3 s

For coming back to initial position where it is thrown , it again takes 3 s , so

Total time of flight = 3 + 3 = 6 s

(c) Position of ball at time t = 5 seconds

Ball takes 3 s to reach max. height and again falls back from there ( from rest )

So we need to find position of ball at 2 s (5-3)

:\implies \sf s=ut+\dfrac{1}{2}at^2:⟹s=ut+

2

1

at

2

:\implies \sf s=(0)(2)+\dfrac{1}{2}(10)(2)^2:⟹s=(0)(2)+

2

1

(10)(2)

2

:\implies \boxed{\sf s=20\ m}:⟹

s=20 m

Distance from the ground = 45 m - 20 m = 25 m