\({v_x} = 25\cos 60^\circ = 12.5\,m/s\)
Vertical component of velocity
\({v_y} = 25\sin 60^\circ = 12.5\sqrt 3 \,m/s\)
Time to cover \(50 \,m\) distance \(t = \frac{{50}}{{12.5}} = 4\,\sec \)
The vertical height y is given by
\(y = {v_y}t - \frac{1}{2}g{t^2} = 12.5\sqrt 3 \times 4 - \frac{1}{2} \times 9.8 \times 16 = 8.2\,m\)
કોલમ $-I$ | કોલમ $-II$ |
$(1)$ કોણીય વેગમાન | $(a)$ અદિશ |
$(2)$ સ્થિતિઊર્જા | $(b)$ સદિશ |
$(c)$ એકમ સદિશ |