A few days ago
Anonymous

Physics hw please help me.?

1.) How much farther will a golf ball with an initial speed of 75 m/s go when projected at 45 degrees than when projected at 30 degrees?

2.) Observations are made of a distant television tower from each end of a 100-m base line. From each position the angle between the line of sight and the base line is 85.5 degrees. How far is the tower from the center of the line?

Top 1 Answers
A few days ago
Anonymous

Favorite Answer

1.

If the ball is projected with velocity v at angle x to the horizontal and t is the time of flight, then:

0 = v sin(x) – gt / 2 …(1)

s = tv cos(x) …(2)

Eliminating t:

s = 2v^2 sin(x)cos(x) / g

s = v^2 sin(2x) / g

If the distances are s1 with x = 30deg, and s2 with x = 45deg, then:

s2 – s1

= v^2 (sin(90) – sin(60)) / g

= 75^2 (1 – sqrt(3)/2)) / 9.81

= 76.82m.

2.

Let A be one end of the base line, C be the base of the tower, B be the mid-point of the baseline.

Triangle ABC:

AB = 50m

BC / AB = tan(85.5)

Hence:

BC = 50tan(85.5) = 635.3m.

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