Solving Trajectory Problem: Initial Speed, Time, & Height

[PhilCam]

Homework Statement

A college regulation goal post has a horizontal ball at a height of 3.048 m from the ground. A kicker attempts a field goal at a distance of 30 m from the goal posts. The initial speed of the ball is 19 m/s making an angle of 40 degrees above the horizontal. The plane of trajectory is perpendicular to the plane of the goal posts. Neglect friction and air resistance.

a) The horizontal component of the intial speed of the ball is:
b) The vertical component of the inital speed of the ball is:
c) The time it takes for the ball to cross the plane of the goal posts is:
d) The ball crosses the plane of the goal posts at a height of:
e) The magnitude of speed of the football as it crosses the plane of the goal posts is:

The Attempt at a Solution

Well I have found (A) using 19m/s X cos 40 degrees and part (B) using 19 m/s X sin 40 degrees.

The answers are 14.55 m/s and 12.2 m/s respectively. For part C, I just took 30 m divided by the horizontal component of 14.55 and the answer was 2.06 seconds which I believe is correct.

For part D and E, I am unsure of how to proceed. I thought maybe 2.06seconds X 12.2 m/s but the answer seemed way too high. Any help would be greatly appreciated.

 

[rl.bhat]

d) Use the formula
y = v*sinθ*t - 0.5*g*t^2. From c) you know the time. Find y.
e) Using v = v*sinθ - g*t, find the vertical component of the velocity after time t. Horizontal component remains constant. Find the resultant of these two velocities.

 

[PhilCam]

Thank you for the help.

Using your second formula:
V=19*(sin 40) - 9.8(2.016)
V= 12.213 - 19.757
V= -7.544

Magnitude = Square root (-7.544^2+14.55^2)
M = Square root (56.912+211.703)
M=16.389

It is telling me the answer should be 16.6.

I have re-done this equation a couple times, is there a math error or an answer key error?

Again, thank you for the help

 

[rl.bhat]

Your math is correct.