TG EAMCET 4 May 2025 Shift 1 Paper

Step 1: Find the total time the object is in the airThe question says the object crosses a certain point after 2.3 s , then takes 5.7 s from that point to reach the ground. So, the total time in the air is:$T=2.3\ \mathrm{s}+5.7\ \mathrm{s}=8\ \mathrm{s}$Step 2: Find the time taken to reach the maximum heightFor a projectile (an object thrown at an angle), the time to go up to the highest point is half of the total time in the air:$t_{\max}=\;\frac{T}{2}=\;\frac{8\ \mathrm{s}}{2}=4\ \mathrm{s}$Step 3: Find the starting upward speed using the motion formulaAt the highest point, the speed going up becomes zero. The formula is:$v_f=v_i-g t_{\max}$Here, $v_f=0$ (since upward speed is zero at the top), $g=10\ \mathrm{m}/\mathrm{s}^2$, and $t_{\max}=4\ \mathrm{s}$ :$0=v_i-(10\times 4) v_i=40\ \mathrm{m}/\mathrm{s}$Step 4: Calculate the maximum height reachedWe use another formula to find the highest point:$H_{\max}=v_i t_{\max}-\;\frac{1}{2} g t_{\max}^2$Plug in the values:$H_{\max}=(40\ \mathrm{m}/\mathrm{s})(4\ \mathrm{s})-\;\frac{1}{2} (10\ \mathrm{m}/\mathrm{s}^2) (4\ \mathrm{s})^2$$\;(40\times 4)=160\ \mathrm{m}$$\;(4\ \mathrm{s})^2=16\ \mathrm{s}^2$$\;\;\frac{1}{2}\times 10\times 16=80\ \mathrm{m}$$\;H_{\max}=160\ \mathrm{m}-80\ \mathrm{m}=80\ \mathrm{m}$Final Answer:The maximum height reached by the body is $80\ \mathrm{m}$.