A body is projected at an angle of 30° with the horizontal and with a speed of 30ms- 1 . What is the angle with the horizontal after 1.5 seconds? (Take g=10ms- 2 )

Question

A body is projected at an angle of 30° with the horizontal and with a speed of 30ms- 1 . What is the angle with the horizontal after 1.5 seconds? (Take g=10ms- 2 ) (A) 0°  (B) 30°  (C) 60°  (D) 90°

Tardigrade Admin · Accepted Answer

The time of flight is given by: T=(2 u sin θ /g)=(2 × 30 × 1/10 × 2)=3sec Thus, after 1.5 sec the body is at the highest point. As the direction of motion is horizontal after 1.5 seconds, the angle with the horizontal is 0° .

Q. A body is projected at an angle of $3 0^{\circ}$ with the horizontal and with a speed of $30 m s^{- 1}$ . What is the angle with the horizontal after $1.5$ seconds? (Take $g = 10 m s^{- 2}$ )

$0^{\circ}$

$3 0^{\circ}$

$6 0^{\circ}$

$9 0^{\circ}$

The time of flight is given by :
$T = \frac{2 u s in θ}{g} = \frac{2 \times 30 \times 1}{10 \times 2} = 3 sec$
Thus, after 1.5 sec the body is at the highest point. As the direction of motion is horizontal after $1.5$ seconds, the angle with the horizontal is $0^{\circ} .$

Q. A body is projected at an angle of 30∘ with the horizontal and with a speed of 30ms−1 . What is the angle with the horizontal after 1.5 seconds? (Take g=10ms−2 )

0∘

30∘

60∘

90∘