A body is projected horizontally from the top of a tower with a velocity of 10 m/s. If it hits the ground at an angle of 45°, then the vertical component of velocity when it hits ground (in m/s) is

Question

A body is projected horizontally from the top of a tower with a velocity of 10 m/s. If it hits the ground at an angle of 45°, then the vertical component of velocity when it hits ground (in m/s) is (A) 10√2 (B) 5√2 (C) 5 (D) 10

Tardigrade Admin · Accepted Answer

tan 45°=(Vy/Vx) 1=(Vy/Vx) V y = V x =10 m / s <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/7a72db114d729939e5f3b4ba50cc1abf-.png" />

Q. A body is projected horizontally from the top of a tower with a velocity of $10 m / s$ . If it hits the ground at an angle of $4 5^{\circ}$ , then the vertical component of velocity when it hits ground (in m/s) is

$102$

$52$

$5$

$10$

$tan 4 5^{\circ} = \frac{V _{y}}{V _{x}}$
$1 = \frac{V _{y}}{V _{x}}$
$V_{y} = V_{x} = 10 m / s$

Q. A body is projected horizontally from the top of a tower with a velocity of 10m/s. If it hits the ground at an angle of 45∘, then the vertical component of velocity when it hits ground (in m/s) is

102​

52​

5

10

tan45∘=Vx​Vy​​ 1=Vx​Vy​​ Vy​=Vx​=10m/s

Q. A body is projected horizontally from the top of a tower with a velocity of $10 m / s$ . If it hits the ground at an angle of $4 5^{\circ}$ , then the vertical component of velocity when it hits ground (in m/s) is

$102$

$52$

$5$

$10$

$tan 4 5^{\circ} = \frac{V _{y}}{V _{x}}$
$1 = \frac{V _{y}}{V _{x}}$
$V_{y} = V_{x} = 10 m / s$