Question

A ball is thrown vertically upwards with a velocity of 10ms−1 from the top of a multi storey building.The height of the point where the ball is thrown 25 m from the ground.how high will the ball raise with respect to time?​

Answer 1

Given:-

• Initial Velocity = 10m/s

• Final Velocity = 0m/s ( As it Covers maximum height )

• Acceleratiom due to gravity = -10m/s ( Upward)

• Height of Building = 25m.

To Find:-

• The Height will it cover after being thrown and time taken to cover that.

Formulae used:-

• v² - u² = 2gh

• S = ut + ½ × a × t²

Where,

• v = Final Velocity
• u = Initial Velocity
• g = Acceleration
• h = Height
• t = Time.

Now,

→ v² - u² = 2gh

→ (0)² - (10)² = 2 × -10 × h

→ -100 = -20h

→ 100 = 20h

→ h = 100/20

→ h = 5m.

Hence, The Distance covered by ball from the building is 5m and from ground it is 30m high.

Now,

• We will find the time which the stone taken to cover the distance.

So,

→ 5 = ut + ½ × a × t²

→ 5 = 10 × t + ½ × -10 × t²

→ 5 = 10t + (-5t²)

→ 5 = 10t - 5t²

→ 5t² - 10t + 5 = 0

→ 5t² -5t - 5t + 5 = 0

→ 5t( t - 1) - 5( t - 1) = 0

→ 5t - 5 = 0 → 5t = 5 → t = 1s

→ t - 1 = 0 → t = 1.

Hence, The Time taken by the stone to cover 5m is 1 second

Answer 2

$\huge\boxed{Answer}$

Time to reach maximum height can be obtained from v=u+at

0=20+(−10)t

t=2s

s=ut+0.5at2=20(2)+0.5(−10)(2)2=20m

Thus, total distance for maximum height is 45 m

s=ut+0.5at2

45=0+0.5(10)(t′)2

t′=3s

Total time= 3+2= 5s

$\huge\mathfrak\red{itz\:jyotsana☺}$