A golfer on a level fairway hits a ball at an angle of 42* to the horizontal that travels 100 yd before striking the ground. He then hits another ball from the same spot with the same speed, but at a different angle. This ball also travels 100 yd. At what angle was the second ball hit? (Neglect air resistance.)
Oh my, I suppose I will use feet and inches and miles and pounds and g = 32 ft/s^2
s = speed in ft/s
u = horizontal speed = s cos 42 = .743 s
Vi = s sin 42 = .669 s
t = time rising
2 t = time in air
300 feet = 2u t = 1.49 s t
v = Vi - 32 t
0 = .669 s - 32 t
t = .0209 s
so
300 = 1.49 s (.0209)s = .03115 s^2
s = 98.1 feet/second
now the second half
u = 98.1 cos T
Vi = 98.1 sin T
t is time rising again
300 feet = 2 u t = 196 cos T * t
0 = 98.1 sin T - 32 t
t = 3.07 sin T
so
300 = 196 cos T * 3.07 sin T
300 = 601 cos T sinT
cos T sin T = .499
2 sin T cos T = .998 = sin 2 T
so
2T = sin^-1 .998 = 86.7 deg
so
T = 43.3 deg
I suspect an arithmetic error because max range is at T = 45 degrees. the result should be as much above 45 as the original was below so It should have come out around 48 degrees.
A golfer on a level fairway hits a ball at an angle of 42* to the horizontal that travels 100 yd before striking the ground. He then hits another ball from the same spot with the same speed, but at a different angle. This ball also travels 100 yd. At what angle was the second ball hit? (Neglect air resistance.)
A golfer on a level fairway hits a ball at an angle of 39° to the horizontal that travels 86 yd before striking the ground. He then hits another ball from the same spot with the same speed, but at a different angle. This ball also travels 86 d. At what angle was the second ball hit? (Neglect air resistance.)
A golfer on a level fairway hits a ball at an angle of 23° to the horizontal that travels 104 yd before striking the ground. He then hits another ball from the same spot with the same speed, but at a different angle. This ball also travels 104 yd. At what angle was the second ball hit? (Neglect air resistance.)
To solve this problem, we can use the concept of projectile motion. We know that the first ball was hit at an angle of 42 degrees and traveled a distance of 100 yards. We need to find the angle at which the second ball was hit, given that it also traveled 100 yards.
Let's break down the motion of the first ball:
1. The horizontal component of the velocity (Vx) remains constant throughout the motion because there is no acceleration acting in that direction.
2. The vertical component of the velocity (Vy) changes due to the acceleration due to gravity.
Now, let's use the equations of projectile motion to find the initial vertical velocity (V0y) and the time of flight (t) for the first ball:
1. The horizontal distance traveled (x) can be calculated as x = Vx * t.
In this case, x = 100 yards.
2. The time of flight can be calculated using the equation t = x / Vx.
3. The horizontal and vertical components of the initial velocity (V0x and V0y) are related to the angle of projection (θ) as follows:
V0x = V * cos(θ)
V0y = V * sin(θ)
Given that the distance traveled (x) and the angle of projection (θ) are known, we can find the initial vertical and horizontal velocities:
V0x = V * cos(42°)
V0y = V * sin(42°)
Since the speed of the second ball is the same as the first ball (V0x and V0y are the same), we can use the equations to find the angle at which the second ball was hit:
V0x = V * cos(θ2)
V0y = V * sin(θ2)
By substituting the known values (V0x and V0y) into the equations above, we can solve for θ2:
V * cos(42°) = V * cos(θ2)
V * sin(42°) = V * sin(θ2)
Dividing both equations by V, we get:
cos(42°) = cos(θ2)
sin(42°) = sin(θ2)
Now, we can use the trigonometric identities to find the values of θ2:
θ2 = cos^(-1)(cos(42°))
θ2 = sin^(-1)(sin(42°))
Calculating these values will give us the angle at which the second ball was hit.