Question

A ball is thrown vertically upwards with a velocity of \(19.6 ms^{–1}\) from the top of a tower. The ball strikes the ground after \(6 s\). The height from the ground up to which the ball can rise will be \((k/5) m\). The value of k is ______ (use \(g = 9.8 m/s^2\))

Answer 1

Given : Initial velocity \(u = 19.6\,m/s\), time \(t = 6\,s\), \(g = 9.8\,m/s^2\)

Let height of tower = \(h\)

Using equation of motion (taking upward positive)

\(\begin{array}{l} s = ut - \frac{1}{2}gt^2 \end{array}\)

Here displacement to ground = \(-h\)

\(\begin{array}{l} -h = 19.6 \times 6 - \frac{1}{2}\times 9.8 \times 6^2 \end{array}\)

\(\begin{array}{l} -h = 117.6 - 176.4 = -58.8 \end{array}\)

\(\begin{array}{l} h = 58.8\,m \end{array}\)

Maximum rise above point of projection

\(\begin{array}{l} h_1 = \frac{u^2}{2g} = \frac{(19.6)^2}{2\times 9.8} \end{array}\)

\(\begin{array}{l} h_1 = 19.6\,m \end{array}\)

Total height from ground

\(\begin{array}{l} H = h + h_1 = 58.8 + 19.6 = 78.4\,m \end{array}\)

\(\begin{array}{l} 78.4 = \frac{k}{5} \end{array}\)

\(\begin{array}{l} k = 78.4 \times 5 = 392 \end{array}\)

Hence, \(k = \mathbf{392}\).