Question

From the top of a tall building, a particle is projected horizontally with a speed of 10 m/s. The speed of the particle at t = 1 s will be

Answer 1

Given :

➾ Initial speed = 10m/s

To Find :

➠ Speed of projectile at t = 1s.

Concept :

➨ This question is completely based on the concept of height to ground projectile motion.

➨ In this motion, Horizontal speed of projectile remains constant throughout the motion because no acceleration acts in horizontal direction.

➨ Vertical speed of projectile changes with time because acceleration due to gravity acts continuously in downward direction.

[Initial vertical speed = zero]

Solution :

$\dashrightarrow\sf\:V=\sqrt{{V_x}^2+{V_y}^2}\\ \\ \dashrightarrow\sf\:V=\sqrt{u^2+(gt)^2}\\ \\ \dashrightarrow\sf\:V=\sqrt{(10)^2+(10\times 1)^2}\\ \\ \dashrightarrow\sf\:V=\sqrt{200}\\ \\ \dashrightarrow\underline{\boxed{\bf{\purple{V=10\sqrt{2}\:ms^{-1}}}}}\:\orange{\bigstar}$

Answer 2

⭐AnswEr⭐

Given:

✴️ initial speed(u)=10 m/s

✴️time(t)=1s

✴️ gravitation (g)=10

SoLutioN:

V=√$V_x$²+$V_y$²

$\implies \: V= \sqrt{ {u}^{2} + {gt}^{2} }$

$\implies \:V= \sqrt{ {10}^{2} + {(10 \times 1)}^{2} }$

$\implies \: V= \sqrt{100 + 100}$

$\implies \: V= \sqrt{200}$

$\implies \: V= \sqrt{100 \times 2}$

$\implies \: V=10 \sqrt{2} \:$

therefore, V=10√2 m/s

⭐ Projectile motion ⭐

When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that is directed towards the centre of the earth (we assume that the particle remains close to the surface of the earth). The path of such a particle is called a projectile and the motion is called projectile motion. Air resistance to the motion of the body is to be assumed absent in projectile motion.

In a Projectile Motion, there are two simultaneous independent rectilinear motions:

♣️Along x-axis: uniform velocity, responsible for the horizontal (forward) motion of the particle.

♣️Along y-axis: uniform acceleration, responsible for the vertical (downwards) motion of the particle.