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A long jump shot is released 1.6 m above the ground, 6.05 m

A long jump shot is released 1.6 m above the ground, 6.05 m

[From: Physics] [author: ] [Date: 01-07] [Hit: ]
A long jump shot is released 1.6 m above the ground, 6.05 m from the base of the basket, which is 3.05 m high. Physics help?For launch angles of 30° and 60°, find the speed needed to make the basket.......

A long jump shot is released 1.6 m above the ground, 6.05 m from the base of the basket, which is 3.05 m high. Physics help?
For launch angles of 30° and 60°, find the speed needed to make the basket.
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electron1 say: The first step is to determine the increase of the height of the shot.
Increase of height = 3.05 – 1.6 = 1.45 meters

The only missing number is the magnitude of the shot’s initial velocity. Let it be v.
Initial vertical velocity = v * sin 30 = v * 0.5
Initial horizontal velocity = v * cos 30 = v * √3/2

During the time the shot is in the air, it has vertical acceleration of -9.8 m/s^2. The horizontal velocity is constant. Let t be the time.

Eq#1: d = v * sin θ * t + ½ * a * t^2
1.45 = v * 0.5 * t – 4.9 * t^2

Eq#2: d = v * cos θ * t
6.05 = v * √3/2 * t
t = 6.05 ÷ v * √3/2

Let’s substitute this for t in the first equation.

1.45 = v * 0.5 * (6.05 ÷ v * √3/2) – 4.9 * (6.05 ÷ v * √3/2)^2
1.45 = 6.05 ÷ √3 – 179.35225 ÷ (v^2 * 0.75)
-6.05 ÷ √3 + 1.45 = – 179.35225 ÷ (v^2 * 0.75
Multiply both sides of this equation by 0.75

-4.5375 ÷ √3 + 1.0875 = – 179.35225 ÷ v^2
v^2 = -179.35225 ÷ (-4.5375 ÷ √3 + 1.0875)
v = √[-179.35225 ÷ (-4.5375 ÷ √3 + 1.0875)]

The magnitude of the initial velocity is approximately 10.7 m/s. Now let’s use the same equations with 60˚.

Initial vertical velocity = v * sin 60 = v * √3/2
Initial horizontal velocity = v * cos 60 = v * 0.5

During the time the shot is in the air, it has vertical acceleration of -9.8 m/s^2. The horizontal velocity is constant. Let t be the time.

Eq#1: d = v * sin θ * t + ½ * a * t^2
1.45 = v * √3/2 * t – 4.9 * t^2

Eq#2: d = v * cos θ * t
6.05 = v * 0.5 * t
t = 12.1 ÷ v

Let’s substitute this for t in the first equation.

1.45 = v * √3/2 * (12.1 ÷ v) – 4.9 * (12.1 ÷ v)2
1.45 = √3 * 6.05 – 717.409 ÷ v^2
Subtract this √3 * 6.05 from both sides of the equation.

1.45 – √3 * 6.05 = -717.409 ÷ v^2
v^2 = -717.409 ÷ (1.45 – √3 * 6.05)
v = √-[717.409 ÷ (1.45 – √3 * 6.05)

The magnitude of the initial velocity is approximately 8.91 m/s. I hope this is helpful for you.
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oubaas say: 30° angle
(3.05-1.6) = Vo*0.5*t-4.90t^2
6.05 = 0.866Vo*t
Vo = 7.0/t
1.45 = 3.5-4.9t^2
t = √2.05/4.9 = 0.65 sec
Vo = 6.05/(0.866*0.65) = 10.8 m/sec

60° angle
(3.05-1.6) = Vo*0.866*t-4.90t^2
6.05 = 0.5Vo*t
Vo = 12.1/t
1.45 = 12.1*0.866-4.9t^2
t = √9.0/4.9 = 1.36 sec
Vo = 6.05/(0.5*1.36) = 8.90 m/sec
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Old Science Guy say: ...
the usual drill is to divide the motion into horizontal (X) motion
and vertical (Y) motion
and these equations
Vx= Vi cosθ
Vy= Vi sinθ
Vx= X/t
Y= Vy*t + 1/2 g t^2
yeah, it gets tedious
so
here's an equation which combines this into one that makes quicker work of most trajectory problems

Y = X*tanθ + ( (g*X^2) / 2(Vi*cosθ)^2 )
if the projectile comes back to ground level, ∆Y = 0 which simplifies things a lot
otherwise higher is positive and lower is negative
X is the downrange horizontal distance
Vi is initial velocity
θ is the launch angle which is usually + but may be 0 or even negative.
all values are in MKS units and g= (-9.81 m/s/s)

in this problem
∆Y= 3.05 -1.6 m = 1.45 m
θ=30° (part 1) 60° (part 2)
Vi= ???
X = 6.05 m
1)
1.45 = 6.05 tan30 + ((-9.81)(6.05^2) / 2(Vi cos30)^2
1.45 = 3.493 - 239.38/Vi^2
239.38 / Vi^2= 20.43
Vi^2 = 11.717
Vi = 3.4 m/s to 2 s.f.

2)
replace 30° with 60° and re-calculate

When you get a good response,
This is the only reward we get.
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