A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground. what is the magnitude of the ball's velocity just before it hits the ground?
a. 9.14 m/s b. 6.79 m/s c. 2.44 m/s d. 1.22 m/s e. 7.45 m/s

Question

A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground. what is the magnitude of the ball's velocity just before it hits the ground?
    a. 9.14 m/s   b. 6.79 m/s   c. 2.44 m/s   d. 1.22 m/s   e. 7.45 m/s

anonymous · Answer

You need to setup a picture and place the motion variable (X,Y,V,T,A) for the beginning and ending pictures 

then....

You can use the vertical position equation 

$$Y_f = Y_i + V_{yi}t_f + \frac{1}{2}a_yt_f$$

where ay = -9.8 m/s^2

anonymous · Answer

whats is the value for yf and yi?

anonymous · Answer

Yi is the intial height from the ground and Yf is the final height from the ground assuming the ground is Y = 0

anonymous · Answer

is yi= 1 sine 40 and yf=0 ?

anonymous · Answer

Yi = 1 m , Yf = 0 m

anonymous · Answer

you will use the sin function to find the components of the velocity to get Vyi

anonymous · Answer

am sorry i wrote the question wrong its supose to ask what is the magnitude of the ball's velocity just before it hits the ground

anonymous · Answer

Ok are you given the total time in the air?

anonymous · Answer

no

anonymous · Answer

the answer is 9.14 m/s but i have no clue how

anonymous · Answer

?