1. In a game of american football, a placekicker mist kick a football from point

Question

1. In a game of american football, a placekicker mist kick a football from point 36.0 . (about 40 yards) from the goal, and half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When he kicked the ball leaves the ground with a speed of 21.0 m/s at the angle 49.0 to the horizontal. By how much does the ball clear or fall short? Enter a negative if it falls short

2. New England Quarterback Tom Brady throws a football straight towards a receiver with an initial speed of 16 m/s at the angle 26 degrees above the horizontal. At that instant, the receiver is 18.5 m from the quarteback. With what constant speed should the receiver run?

3. The determined coyote is out once more in pursiut of the elusive roadrunner. The coyote wears a pair of Acme jet-powered roller skates, which provide a constant horizontal acceleration of 12.0 m/sec^2. The coyote starts at rest 50.0 m from the brink of the cliff at the instant the roadrunner zips past him in the direction of the cliff.

a) minimum speed the roadrunner must have to reach the cliff before the coyote?

b) at the edge of the cliff the roadrunner escapes by making a sudden turn, while the coyote continues straight ahead. The coyote's skaes remain horizontal and continue to operate while he is in flight so that his acceleration while in the air is (12.o i-9.8j) m/s^2. The cliff is 100 m above the flat floor of a wide canyon. Determine where the coyote lands in the canyon

c) determine he components of the coyotes impact velocity (assume right is possitive x direction and up is the possitive y direction)

Explanation / Answer

1) Given d=36.0m, y=3.05m, =49.0, vo=21.0 m/s
Assuming no air resistance, the equations are:
x(t) = vcos()t
y(t) = vsin()t-(g/2)t²

Find time when y(t)=3.05 by using quadratic equation:
(g/2)t²-vsin()t+y(t) = 0
a=g/2, b=-vsin(), c=y
t = (-b±(b²-4ac))/2a
t = (vsin+(v²sin²()-2gy))/g

Plugging this into x(t) above gives:
x(t) = vcos(vsin+(v²sin²()-2gy))/g   
x(t) = 41.73 m
So ball clears crossbar by 41.73 - 36 = 5.73 m

3) a) Time taken for coyote to reach the cliff .
50 = ½12t²
t = 2.9 s
Distance = Speed x Time
50 = Speed x 2.9
Speed = 17.24 m/s .

b) Calculate time taken for its fall .
100 = ½(9.80)t²
t = 4.52 s
Distance travelled horizontally = ½at² + ut
u :-
u² = 2(12)(50)
u = 34.64
x = ½(12)(4.52)² + 34.64*(4.52)
x = 279.2 m  

c) vertically, he falls 100m from rest under gravity
s = 1/2 ft^2 and v = gt eliminating t, we have v^2 = 2gs
so v = sqrt 2 * 9.8 * 100 = 44.27

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