Question 5: Motion in 2D & Vectors

Question

A boy throws a ball into air at $45^{\circ}$ from the horizontal to land it on a roof of a building of height $H$. If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then value of $H$ is $\_\_\_\_$ m.
$$ \left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right) $$

Accepted Answer

Let u be the initial velocity and $	heta=45^{\circ}$ be the angle of projection.
The initial vertical velocity is $\mathrm{u}_{\mathrm{y}}=\mathrm{u} \sin 	heta$
The ball attains maximum height in $\mathrm{t}_{\max }=2 \mathrm{~s}$.
At the maximum height, the vertical component of the velocity becomes zero $\left(\mathrm{v}_{\mathrm{y}}=0ight)$.
Using the first equation of motion $(\mathrm{v}=\mathrm{u}+\mathrm{at})$ :
$$ \mathrm{v}_{\mathrm{y}}=\mathrm{u}_{\mathrm{y}}-\mathrm{gt} $$
$$\Rightarrow $$ $$ 0=\mathrm{u}_{\mathrm{y}}-(10)(2) \Rightarrow \mathrm{u}_{\mathrm{y}}=20 \mathrm{~m} / \mathrm{s} $$
So, the initial vertical velocity is $20 \mathrm{~m} / \mathrm{s}$.
The ball lands on the roof of the building at $\mathrm{t}=3 \mathrm{~s}$.
This means at $t=3$, the vertical displacement $(y)$ of the ball is equal to the height of the building $(H)$.
Using the second equation of motion $\left(s=u t+\frac{1}{2} a t^2ight)$ :
$$ \mathrm{y}=\mathrm{u}_{\mathrm{y}} \mathrm{t}-\frac{1}{2} \mathrm{gt}^2 $$
Substituting the given values :
1. $y=H$ (Height of the building)
2. $\mathrm{u}_{\mathrm{y}}=20 \mathrm{~m} / \mathrm{s}$ (Calculated in Step 2)
3. $\mathrm{t}=3 \mathrm{~s}$ (Given time of landing)
$$ H=(20)(3)-\frac{1}{2}(10)(3)^2=15 \mathrm{~m} $$
Therefore, the height of the building is 15 m. Hence, the correct option is (D).

Answer

20

Answer

25

Answer

10

Answer

15

Motion in 2D & Vectors

A boy throws a ball into air at 45^{\circ} from the horizontal to land it on a roof of a building of height H. If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then value of H is \_\_\_\_ m. $ \left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)

Choose the correct answer: