Question

A ball is thrown vertically upward and it reaches a height of 90 m . Find (a) the velocity with which it was thrown and (b) the height reached by the ball 7 seconds after it was thrown (g=9.8 ms-2)

Answer 1

Given :-

◉ A ball is thrown vertically upward and it reaches a maximum height of 90 m.

⇒ g = 9.8 m/s²

To Find :-

◉ The velocity with which it was thrown

◉ The height reached by the ball after 7s.

Solution :-

It is given that the maximum height achieved by the ball is 90 m from initial velocity u ,

⇒ Maximum Height = u² / 2g

⇒ 90 × 2g = u²

⇒ 180 × 9.8 = u²

⇒ u² = 1764

⇒ u = 42 m/s

Hence, The velocity at which the ball was thrown was 42 m/s.

Now, We need to find the height achieved by the ball 7s after the ball was thrown,

• u = 42 m/s
• t = 7 s
• g = - 9.8m/s² [ g is acting in opposite direction of motion ]

⇒ h = ut + 1/2 gt²

⇒ h = 42 × 7 - 1/2 × 9.8 × 49

⇒ h = 294 - 240.1

⇒ h = 53.9 m

Hence, The height achieved by the ball 7s after it was thrown was 53.9 m.

Answer 2

Use v^2 = Uy^2 + 2gs ( basic equation derived from linear motion eq, you can read up on it online if you want a full derivation.)

Where v is final velocity, Uy is initial velocity in the vertical direction, g is gravity and s is displacement. Assume upwards is +ve and downwards is -ve. Also I am assuming there is no air resistance involved.

At max height; s = 90m, v = 0 and obviously g is always -9.81 m/s^2 unless specified.

Use these constraints in the original equation and re - arrange for u:

Uy = (v^2 - 2gs)^0.5

Uy = (0 - 2×(-9.81)(90))^0.5

Uy = 42.02 m/s upwards

Now if you are given an angle aswell the same problem can be solved in a complex scenario i.e if there is a horizontal component for initial velocity.

Uy is the upward component of the velocity. Now you can use v = Uy + gt to find time taken for the ball to reach this height.

t = (v - Uy)/g

t = (0 - 42.02)/-9.81

t = 4.284 seconds

This time is for the ball to reach max height, therefore assuming a peferctly parabolic path this would give t = 8.567 seconds.

Now use S = Uyt + 0.5gt^2 and t = s / Ux, where Ux = u cos (¥°).

Remember u sin(¥°) is Uy, therefore;

Make the substitutions at S = 0

0 = Uy (s/Ux) + 0.5g(t)^2

0 = s×(sin (¥)/cos (¥) ) + 0.5 (-9.81)(8.567)^2

0 = s×tan (¥) - 360

s = 360/tan (¥)

This is the horizontal range of the projectile. Using this and Ux = s/t will give you horizontal component of initial velocity.