Solution:
Given,
Initial velocity (u) = 72 km/hr,
= \(\frac{72\times1000}{60\times60}\) m/s
= \(\frac{720}{36}\) m/s
= 20 m/s
Final velocity (v) = 0 m/s [ because the body is thrown vertically upward ]
Acceleration (a) = - g = -10 m/s2
Height reached (h) = ?
TIme taken (t) = ?
we have from the third equation of motion in straight line..
\(v^2=u^2+2as\)
Or, \(0^2=20^2+2\times(-10)\times h\) [ we consider s = h for object falling vertically upward or downward ]
Or, 0 = 400- 20h
Or, 20h = 400
Or, h = 400/ 20
∴ h = 20 m
Also, from the first equation of motion in a straight line, we have
v = u + at
Or, 0 = 20 + (-10)t
Or, -20 = -10t
∴ t = 2 s
Hence, the body reaches the height of 20 m in 2 seconds.