Question

A tennis ball is lobbed (from very close to the ground) at a speed of 11.5 m/s and an angle of 47.9 degrees to the horizontal. (a) Find the horizontal distance traveled by the ball (the range) (b) At what other angle will one obtain the same range? (c) Which angle gives the highest point attained by the ball and what is this maximum height? (In all cases ignore air resistance)

Answer 1

Part A.

Range is given by:

R = V0x*T

V0x = Initial horizontal velocity = V0*cos A

T = time period of projectile = 2*V0y/g = 2*V0*sin A/g

So,

R = (V0*cos A)*(2*V0*sin A)/g = (V0^2*sin 2A)/g

Using given values:

V0 = 11.5 m/sec & A = 47.9 deg

g = 9.81 m/sec^2

So,

R = 11.5^2*(sin 2*47.9 deg)/9.81

R = 11.5^2*(sin 95.8 deg)/9.81

R = 13.4 m

Part B.

If projectile motion is thrown at an angle A, then range of projectile will be same at angle (90 - A)

Given that A = 47.9 deg

So at 90 - A = 90 - 47.9 = 42.1 deg range will be same.

to check this

R = 11.5^2*(sin 2*42.1 deg)/9.81

R = 11.5^2*(sin 84.2 deg)/9.81

R = 13.4 m

Part C.

Height of projectile is given by:

H = V0^2*sin²A/(2*g)

So when launching angle is A = 47.9 deg

H1 = 11.5^2*(sin 47.9 deg)^2/(2*9.81) = 3.71 m

And when launching angle is A = 42.1 deg

H2 = 11.5^2*(sin 42.1 deg)^2/(2*9.81) = 3.03 m

So Angle 47.9 deg gives the highest point attained by the ball and this point max height is 3.71 m

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