Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

You kick a soccer ball with a speed of 31 m/s at an angle of 50°. How long does it take the ball to reach the top of its trajectory?

Home/English/Physics/You kick a soccer ball with a speed of 31 m/s at an angle of 50°. How long does it take the ball to reach the top of its trajectory?

You kick a soccer ball with a speed of 31 m/s at an angle of 50°. How long does it take the ball to reach the top of its trajectory?

Question

You kick a soccer ball with a speed of 31 m/s at an angle of 50°. How long
does it take the ball to reach the top of its trajectory?

All credit goes to the person who answered this years ago:) Reaches max height at t = 2.42s. Explanation: I’ve assumed we are neglecting air resistance. If not let me know and I’ll update. We want to examine the behaviour of the ball in the y-direction. In the absence of air resistance the only force acting on the ball is gravity, which produces an acceleration in the negative y direction. From Newton’s 2nd law: m d 2 y d t 2 = − m g m d 2 y d t 2 = − m g Integrating: d y d t = − ∫ g d t d y d t = − g ⋅ t + C From initial conditions, d y d t = v y ( t ) , v 0 = v ( 0 ) = 31 ⋅ sin ( 50 ) ∴ v y ( t ) = v 0 − g ⋅ t The maximum height will be reached at v y = 0 so we solve for t. v y ( t h max ) = 0 ⇒ v 0 = g ⋅ t h max t h max = v 0 g = 31 ⋅ sin ( 50 )

## Answers ( )

Answer:2.42

Explanation:All credit goes to the person who answered this years ago:)

Reaches max height at t = 2.42s.

Explanation:

I’ve assumed we are neglecting air resistance. If not let me know and I’ll update.

We want to examine the behaviour of the ball in the y-direction. In the absence of air resistance the only force acting on the ball is gravity, which produces an acceleration in the negative y direction. From Newton’s 2nd law:

m

d

2

y

d

t

2

=

−

m

g

m

d

2

y

d

t

2

=

−

m

g

Integrating:

d

y

d

t

=

−

∫

g

d

t

d

y

d

t

=

−

g

⋅

t

+

C

From initial conditions,

d

y

d

t

=

v

y

(

t

)

,

v

0

=

v

(

0

)

=

31

⋅

sin

(

50

)

∴

v

y

(

t

)

=

v

0

−

g

⋅

t

The maximum height will be reached at

v

y

=

0

so we solve for t.

v

y

(

t

h

max

)

=

0

⇒

v

0

=

g

⋅

t

h

max

t

h

max

=

v

0

g

=

31

⋅

sin

(

50

)