Kinematics long jumper problem

[kmikias]

Hi ...I came up another question again i guess you will help me out .Here is the question

1. A long jumper takes off at an angle of 20 degree with the horizontal and reaches a maximum height of 0.55 m at mid-flight.
- how long is she in the air?
- what is the forward component of her velocity?
- how far does she jump?

Here is what i try.

A. how long is she in the air.
i use the formula Y= vt + 1/2gt^2
which is -0.55 = 1/2 * -9.8 t^
so t= 0.335 sec and finaly i multiplay 0.335 * 2 = 0.67 sec.
B. here is where i stop how can i find her velocity without knowing the horizontal distance.

i know the formula X = vt

thank you.

 

[LowlyPion]

Hi ...I came up another question again i guess you will help me out .Here is the question

1. A long jumper takes off at an angle of 20 degree with the horizontal and reaches a maximum height of 0.55 m at mid-flight.
- how long is she in the air?
- what is the forward component of her velocity?
- how far does she jump?

Here is what i try.

A. how long is she in the air.
i use the formula Y= vt + 1/2gt^2
which is -0.55 = 1/2 * -9.8 t^
so t= 0.335 sec and finaly i multiplay 0.335 * 2 = 0.67 sec.
B. here is where i stop how can i find her velocity without knowing the horizontal distance.

i know the formula X = vt

thank you.

Think about what the horizontal and vertical components are of her initial velocity.
(Hint: They give you an angle.)

Horizontal velocity has nothing to slow it so you are right that distance will be Vx * t.
Vertical distance is given so you know the total time from doubling how long it would take for her to fall. And that you know as = 1/2 g*t²

 

[kmikias]

yes i did but i still didn't get that because how can i find with only one given angle

 

[LowlyPion]

yes i did but i still didn't get that because how can i find with only one given angle

That's how you solve the problem.

V_y = V*Sinθ is the Y component of the initial velocity.
V_x = V*Cosθ is the horizontal component.