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a golf ball is hit with the velocity of 24.5 m/s at 35 degrees above the horizantle. find:range and max height?
Okay, to help you understand this, begin by drawing a right angled triangle, where the hypotenuse is 24.5 m/s and the angle θ (is the angle between the hypotenuse and the horizontal component. Let θ = 35°.
Then using trigonometry, find the vertical component of velocity (OPPOSITE) and the horizontal component of velocity (ADJACENT)
u(y) = 24.5sin35 = 14.05 m/s
u(x) = 24.5cos35 = 20.07 m/s
Okay, begin by finding the max. height. The ball is at the maximum height when the velocity is 0. Becuase when you hit the ball, it starts at its initial velocity, slows down, comes to a halt (at the max. height) and then drops back down.
We'll use this equation to find the max height.
v² = u² + 2as (note that u, refers to the initial vertical component velocity)
0² = (14.05)² + 2(-9.8)s
197.4 = 19.6s
Therefore s = 10.07 metres. Therefore the max height is approximately 10m.
Now in order to find the range I am going to find the total time taken. Begin by finding the time taken to reach the maximum height.
We will use this equation to find the time taken to reach the max. height. REMEMBER: At the max. height, v = 0
v = u + at (u refers to vertical velocity component)
0 = 14.05 + (-9.8)t
14.05 = 9.8 t
Therefore t = 1.43 s
NOTE: THAT IS THE TIME TAKEN TO REACH THE MAX HEIGHT. SO THEREFORE, THE TOTAL TRIP TIME MUST BE DOUBLE THAT AS IT TAKES THE SAME TIME TO GET BACK DOWN.
Therefore, total trip time = 2 x 1.43
= 2.86 s
Since the horizontal initial velocity is a constant we can use the formula, vt = s to find the range (where s is the range)
SO: s = vt
s = 2.86 x 20.07
s = 57.5 m (RANGE)
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