Question:
A ball is thrown upward from the top of a building at an angle of \(30^\circ\) to the horizontal and with an initial speed of \(20\ \text{m s}^{-1}\). If the ball strikes the ground after \(3\) s, then the height of the building is
\[
(\text{acceleration due to gravity }=10\ \text{m s}^{-2})
\]
A ball is thrown upward from the top of a building at an angle of \(30^\circ\) to the horizontal and with an initial speed of \(20\ \text{m s}^{-1}\). If the ball strikes the ground after \(3\) s, then the height of the building is
\[ (\text{acceleration due to gravity }=10\ \text{m s}^{-2}) \]
\[ (\text{acceleration due to gravity }=10\ \text{m s}^{-2}) \]
Show Hint
In projectile motion, always resolve the initial velocity into horizontal and vertical components first:
\[
u_x=u\cos\theta,\qquad u_y=u\sin\theta.
\]
Then apply the vertical motion equation
\[
s=u_y t-\frac{1}{2}gt^2
\]
to determine heights and vertical displacements.
Updated On: Jun 26, 2026
- \(10\ \text{m}\)
- \(15\ \text{m}\)
- \(20\ \text{m}\)
- \(25\ \text{m}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Resolve the initial velocity into components.
The ball is projected with speed \[ u=20\ \text{m s}^{-1} \] at an angle \[ 30^\circ \] to the horizontal.
The vertical component of the initial velocity is \[ u_y=u\sin30^\circ \] \[ u_y=20\times \frac{1}{2} \] \[ u_y=10\ \text{m s}^{-1}. \]
Step 2: Take upward direction as positive.
Let the height of the building be \(h\).
The ball reaches the ground after \[ t=3\ \text{s}. \] The vertical displacement from the point of projection to the ground is \[ s=-h. \] Using the equation \[ s=u_y t+\frac{1}{2}at^2, \] where \[ a=-g=-10\ \text{m s}^{-2}, \] we get \[ -h=(10)(3)+\frac{1}{2}(-10)(3)^2. \]
Step 3: Simplify the equation.
\[ -h=30-5(9). \] \[ -h=30-45. \] \[ -h=-15. \] Therefore, \[ h=15. \]
Step 4: Verify the result.
The ball initially rises due to the upward component of velocity and then falls under gravity. The net vertical displacement after \(3\) s is \[ -15\ \text{m}, \] which means the ground is \(15\) m below the point of projection.
Hence the building height is correctly obtained as \(15\) m.
Step 5: Final conclusion.
Therefore, the height of the building is \[ \boxed{15\ \text{m}} \] Hence, the correct option is \[ \boxed{(2)} \]
The ball is projected with speed \[ u=20\ \text{m s}^{-1} \] at an angle \[ 30^\circ \] to the horizontal.
The vertical component of the initial velocity is \[ u_y=u\sin30^\circ \] \[ u_y=20\times \frac{1}{2} \] \[ u_y=10\ \text{m s}^{-1}. \]
Step 2: Take upward direction as positive.
Let the height of the building be \(h\).
The ball reaches the ground after \[ t=3\ \text{s}. \] The vertical displacement from the point of projection to the ground is \[ s=-h. \] Using the equation \[ s=u_y t+\frac{1}{2}at^2, \] where \[ a=-g=-10\ \text{m s}^{-2}, \] we get \[ -h=(10)(3)+\frac{1}{2}(-10)(3)^2. \]
Step 3: Simplify the equation.
\[ -h=30-5(9). \] \[ -h=30-45. \] \[ -h=-15. \] Therefore, \[ h=15. \]
Step 4: Verify the result.
The ball initially rises due to the upward component of velocity and then falls under gravity. The net vertical displacement after \(3\) s is \[ -15\ \text{m}, \] which means the ground is \(15\) m below the point of projection.
Hence the building height is correctly obtained as \(15\) m.
Step 5: Final conclusion.
Therefore, the height of the building is \[ \boxed{15\ \text{m}} \] Hence, the correct option is \[ \boxed{(2)} \]
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