A particle is projected vertically upwards with a speed of 40 m/s. At what times will it be 35m above its point of projection? \(\text{Take g} = 10 ms^{-2}\)
\(s = ut + \frac{1}{2}at^{2}\)
\(s = ut - \frac{1}{2}gt^{2}\) (Upward movement against gravity)
\(35 = 40t - \frac{1}{2}10t^{2}\)
\(35 = 40t - 5t^{2}\)
\(5t^{2} - 40t + 35 = 0\)
\(t^{2} - 8t + 7 = 0\)
\((t - 1)(t - 7) = 0 \implies t = \text{1 sec and 7 sec}\)
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}