nineeyes
- 21
- 0
A football reciever B runs the slant in pattern, making a cut at P and thereafter running with a constant speed v_B=21ft/sec at an angle of 30 degrees from the y-axis. The quarterback releases the ball with a Horizontal velocity v_A_x=100ft/sec the instant the reciever passes pont P. Determine the angle at \alpha (the angle between Va and the x-axis) and determine the velocity of the ball relative to the reciever when the ball is caught. Neglect any vertical motion of the ball.
The answer is given \alpha=33.3 degrees and v_A_/_B= (73.1 i + 73.1 j)ft/sec
I tried to solve this by finding the time both the football and the reciever reach the same distance by using 100*t = 21*sin(30)*t+45
I used this time to find the displacement of the reciever from (45,45)... then I tried to use tan(\frac{s_y}{s_x})=\alpha. (thes_y and s_x I used were s_y=35.85 and s_x=50.28But the alpha I get is 35.488 degrees, which is quite a bit higher. I was wondering what I am doing wrong.
Thanks in advance
The answer is given \alpha=33.3 degrees and v_A_/_B= (73.1 i + 73.1 j)ft/sec
I tried to solve this by finding the time both the football and the reciever reach the same distance by using 100*t = 21*sin(30)*t+45
I used this time to find the displacement of the reciever from (45,45)... then I tried to use tan(\frac{s_y}{s_x})=\alpha. (thes_y and s_x I used were s_y=35.85 and s_x=50.28But the alpha I get is 35.488 degrees, which is quite a bit higher. I was wondering what I am doing wrong.
Thanks in advance
Attachments
Last edited: