Question

A body is thrown with a velocity of 9.8m/s making an angle of 30° with horizontal .It will hit the ground after a time

Answer 1

Answer:-

T = 1s

Given:-

u = 9.8 m/s

$\theta = 30^{\circ}$

g = 9.8 m/s²

To find :-

The time taken by body to reach the ground.

Solution:-

Projectile motion :- When a body is projected obliquely from earth surface with certain angle then the body cover a parabolic path and strike with ground.

Motion of the such body is called projectile motion.

As per equation of the projectile motion,

We have ,

$T = \dfrac{2u Sin\theta }{g}$

→$T = \dfrac{2 \times 9.8 Sin30^{\circ}}{9.8}$

→$T = \dfrac{19.6 \times \dfrac{1}{2}}{9.8}$

→$T = \dfrac{19.6}{9.8 \times 2}$

→$T = \dfrac{19.6}{19.6}$

→$T = 1s$

hence,

Time taken by body to hit the group is 1 s.

Answer 2

Answer:

The time taken by the ball to reach the ground = 1 second

Explanation:

Given that,

A body is thrown with a velocity of 9.8m/s making an angle of 30°

and

we know that,

when a body is thrown at an angle then it is the case of projectile motion,

Projectile motion -> When an object or particle is thrown near the Earth's surface and

moves along a curved path at a certain angle

under the action of gravity only

and cover a parabolic path

then such a type of motion is called parabolic motion.

here,

given the initial velocity of the particle = 9.8 m/s

and angle at which it was thrown = 30°

and we know that,

in this case

time taken to reach the ground =

2u sinф/g

here,

u = initial velocity = 9.8 m/s

ф = 30°

g = gravitational acceleration

= 9.8 m/s²

putting the values,

time taken = 2 × 9.8 × sin30°/9.8

2 × 1/2

= 1 s

so,

The time taken by the ball to reach the ground

= 1 second