A football is kicked with an initial velocity of 64ft/s at a projectile angle of 45degrees...?

Question

A receiver 60yd away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to catch the ball just before it hits the ground? Neglect air Resistance.

Answer 1

I can't convert Imperial measurements, but I can explain the theory.

Any vector can be broken down into two or more component vectors. Your ball has a momentum (mv) from the impulse provided in the act of kicking.

Velocity has a vector and you have been given it, 45 degrees at 64ft/s. Hence you have the Earth and your ball traveling relative to it. Gravity is going to act on this ball pulling it to the ground (technically the Earth is curved but we will consider it as flat because the trajectory is going to be a small perabolla).

So we can say that while the ball is in flight it's weight is going to pull it back to Earth. Considering that at the apex of the flight the ball is no longer going to be going up a vertical velocity of zero we can consider the simple relationship

gt + u = v

u will be zero, g is the acceleration due to gravity (9.81m/s/s Note: your constant will be different if working in ft/s) and v will be the component of vertical velocity it initially had (conservation of energy, by neglecting resistive forces).

The vertical velocity will be root(32 f/s) (from Pythagoras) or 64sin 45 ft/s (from trigonometry).

Hence you have found half the time the ball was in the air. Double this for the total time in the air.

The horizontal velocity of the ball will be 64cos 45 ft/s from here is is d=vt (horizontal velocity) to give where the ball will land. Subtract from that 60 yards in feet giving you a new distance K

Understanding K = the receivers average speed X time in the air

The receivers speed will be give by dividing K by the time the ball spent in the air.

Answer 2

The range (maximum horizontal distance) travelled by a projectile is given by the formula

R = V^2(sin 2A)/g

where

R = range

V = initial velocity of ball = 64 ft/sec. (given)

A = angle of launch = 45 degrees (given)

g = acceleration due to gravity = 32.2 ft/sec^2 (constant)

Substituting values,

R = 64^2(sin 2*45)/9.8

R = 127.20 feet

Since the receiver is 60 yards away (180 feet), he will have to travel a distance of 180 - 127.20 = 52.80 feet to catch the ball.

The ball's total travel time is given by the formula

T = 2V(sin A)/g

where all the terms have been previously defined.

Substituting values,

T = 2(64)(sin 45)/32.2

T = 2.82 sec.

Therefore, in order for the receiver to catch the ball, his speed must be equal to

52.80/2.82 = 18.72 ft/sec.

Hope this helps.