TG EAMCET 4 May 2025 Shift 1 Paper

Step 1: Find the total time the object is in the air
The 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=2.3s+5.7s=8s
Step 2: Find the time taken to reach the maximum height
For 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:
tmax=‌

T

2

=‌

8s

2

=4s
Step 3: Find the starting upward speed using the motion formula
At the highest point, the speed going up becomes zero. The formula is:
vf=vi−gtmax
Here, vf=0 (since upward speed is zero at the top), g=10m∕s2, and tmax=4s :
0=vi−(10×4)vi=40m∕s
Step 4: Calculate the maximum height reached
We use another formula to find the highest point:
Hmax=vitmax−‌

1

2

gtmax2
Plug in the values:
Hmax=(40m∕s)(4s)−‌

1

2

(10m∕s2)(4s)2
‌(40×4)=160m
‌(4s)2=16s2
‌‌

1

2

×10×16=80m
‌Hmax=160m−80m=80m
Final Answer:The maximum height reached by the body is 80m.