1
TG EAPCET 2025 (Online) 4th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

A body projected at certain angle $\left(\neq 90^{\circ}\right)$ from the ground crosses a point in its path at a time of 2.3 s and from there it reaches the ground after a time of 5.7 s . The maximum heigh reached by the body is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

A

80 m

B

120 m

C

40 m

D

160 m

2
TG EAPCET 2025 (Online) 3rd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

A ball projected at an angle of $45^{\circ}$ with the horizontal crosses two points at equal heights separated by a distance at times 2 s and 8 s respectively. The horizontal distance between the two points is

(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

A

300 m

B

400 m

C

500 m

D

600 m

3
TG EAPCET 2025 (Online) 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If a body projected with a velocity of $19.6 \mathrm{~ms}^{-1}$ reaches a maximum height of 9.8 m , then the range of the projectile is

(Neglect air resistance)

A

19.6 m

B

78.4 m

C

39.2 m

D

9.8 m

4
TG EAPCET 2025 (Online) 2nd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Two bodies are projected from the same point with the same initial velocity ' $u$ ' making angles ' $\theta^{\prime}$ and $\left(90^{\circ}-\theta\right)$ with the horizontal in opposite directions. The horizontal distance between their positions when the bodies are at their maximum heights is

A

$\frac{u^2}{2 g}\left(\sin ^2 \theta-\cos ^2 \theta\right)$

B

$\frac{u^2 \sin 2 \theta}{2 g}$

C

$\frac{u^2}{g}$

D

$\frac{u^2 \sin 2\left(90^{\circ}-\theta\right)}{g}$

TS EAMCET Subjects

Browse all chapters by subject