A particle is projected with a velocity \(60\text{ }\text{m/s}\) at an angle \(3{0}^{∘}\) with respect to the horizontal…

Explanation

1. Initial Vertical Velocity (\({u}_{y}\)):

   \({u}_{y}=u\sin ⁡\theta =60\sin ⁡3{0}^{∘}=30\text{ }\text{m/s}\mathrm{.}\)

2. Time of Flight (Total time of upward motion):
   Using the formula for the total time of motion upward:

   \({T}_{\text{up}}=\frac{{u}_{y}}{g}=\frac{30}{10}=3\text{ }\text{seconds}\mathrm{.}\)

   The last second in the upward motion corresponds to the time \(t=2\text{ }\text{seconds}\to 3\text{ }\text{seconds}\)

Height in the First Second (\({h}_{1}\)​):

Using the equation of motion:

\(h={u}_{y}t−\frac{1}{2}g{t}^{2},\)

for \(t=1\):

\({h}_{1}=(30)(1)−\frac{1}{2}(10)(1{)}^{2}=30−5=25\text{ }\text{m}\mathrm{.}\)

Height in the Last Second Upward (\({h}_{2}\)):

For the last second in the upward motion, the time interval is \(t=2\to 3\text{ }\text{seconds}\). To find \({h}_{2}\), calculate the difference in height between these two time intervals:

\({h}_{2}=\text{Height at }t=3\text{ }\text{seconds}−\text{Height at }t=2\text{ }\text{seconds}\mathrm{.}\)

Height at \(t=2\):

\(h(t=2)=(30)(2)−\frac{1}{2}(10)(2{)}^{2}=60−20=40\text{ }\text{m}\mathrm{.}\)

Height at \(t=3\):

\(h(t=3)=(30)(3)−\frac{1}{2}(10)(3{)}^{2}=90−45=45\text{ }\text{m}\mathrm{.}\)

Height in the last second:

\({h}_{2}=h(t=3)−h(t=2)=45−40=5\text{ }\text{m}\mathrm{.}\)

Ratio \(\frac{{h}_{1}}{{h}_{2}}\)​​:

\(\frac{{h}_{1}}{{h}_{2}}=\frac{25}{5}=5:1.\)

Final Answer:

\(5:1\)​