In 1998, jason elam kicked a record field goal. the football started on the ground a horizontal distance of 63.0 yards from the goal posts and just barely cleared the 10-ft-high bar. if the initial trajectory of the football was 40.0° above the horizontal, (a) what was its initial speed and (b) how long after the ball was struck did it pass through the goal posts?

a)The initial speed of the ball was 24.8 m/s.

b)The ball passed through the goal posts 3.04 s after it was kicked.

Explanation:

The position vector of the ball at time "t" can be calculated using the following equation:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a better understanding of the problem ( the following unit conversions were made 63.0 yards = 57.6 m and 10 ft = 3.05 m).

Notice that the position vector when the ball clears the bar is the following (see r 1 in the figure) :

r 1 = (57.6 m, 3.05 m)

Then, using the equation for the x and y-component of the position vector written above:

57.6 m = x0 + v0 · t · cos α

3.05 m = y0 + v0 · t · sin α + 1/2 · g · t²

Placing the origin of the frame of reference at the kicking point, x0  and y0 = 0. Then:

57.6 m = v0 · t · cos α

3.05 m = v0 · t · sin α + 1/2 · g · t²

We have two equations and two unknowns, so, we can solve the system:

Let´s solve the equation of the x-component for v0:

57.6 m = v0 · t · cos α

57.6 m / t · cos α = v0  

Now, replacing v0 in the equation of the y-component of r

3.05 m = v0 · t · sin α + 1/2 · g · t²

3.05 m = (57.6 m / t · cos α) · t · sin α + 1/2 · g · t²

3.05 m = 57.6 m · tan 40.0° - 1/2 · 9.8 m/s² · t²

Solving for "t":

(3.05 m - 57.6 m · tan 40.0°) / -  1/2 · 9.8 m/s² = t²

t = 3.04 s

b) The ball passed through the goal posts 3.04 s after it was kicked.

a) The initial velocity can now be calculated:

57.6 m / t · cos α = v0  

v0 = 57.6 m / 3.03 s · cos 40.0° = 24.8 m/s

The initial speed of the ball was 24.8 m/s

weight of the box is given as

here this is placed on the inclined of angle 42 degree

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so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards