Question

Body is thrown from a point with a speed 50 metre per second at an angle 37 degree with horizontal when it has moved a horizontal distance of 80 metre then its distance from the projection

Answer 1

A body is thrown from a point with speed 50m/s at am angle 37° with horizontal.

a/c to question, body moves 80m in horizontal direction. let's find how much time taken to move 80m in horizontal direction.

so, initial velocity of body along horizontal direction is , $u_x=50cos37^{\circ}$

we know, cos37° = 4/5

so, ux = 50 × 4/5 = 40m/s

and acceleration along horizontal direction , ax = 0

now using formula, $x=u_xt+\frac{1}{2}a_xt^2$

or, 80 = 40t => t = 2sec

now find distance travelled during 2 sec in vertical direction.

initial velocity of body along vertical direction, [u_y=50sin37^{\circ}[/tex]

= 50 × 3/5 = 30 [as sin37° = 3/5 ]

now using formula, $y=u_yt+\frac{1}{2}a_yt^2$

= 30 × 2 + 1/2 × (-10) × (2)²

= 60 - 20

= 40m

hence, distance travelled by body in vertical direction is 40m

Answer 2

Answer:

hyy mate here is ur answer

Explanation:

A body is thrown from a point with speed 50m/s at am angle 37° with horizontal.

a/c to question, body moves 80m in horizontal direction. let's find how much time taken to move 80m in horizontal direction.

so, initial velocity of body along horizontal direction is , u_x=50cos37^{\circ}u

x

=50cos37

∘

we know, cos37° = 4/5

so, ux = 50 × 4/5 = 40m/s

and acceleration along horizontal direction , ax = 0

now using formula, x=u_xt+\frac{1}{2}a_xt^2x=u

x

t+

2

1

a

x

t

2

or, 80 = 40t => t = 2sec

now find distance travelled during 2 sec in vertical direction.

initial velocity of body along vertical direction, [u_y=50sin37^{\circ}[/tex]

= 50 × 3/5 = 30 [as sin37° = 3/5 ]

now using formula, y=u_yt+\frac{1}{2}a_yt^2y=u

y

t+

2

1

a

y

t

2

= 30 × 2 + 1/2 × (-10) × (2)²

= 60 - 20

= 40m

hence, distance travelled by body in vertical direction is 40m

hope it's helpful

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