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.

answers:
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.

oubaas say: 30° angle
(3.051.6) = Vo*0.5*t4.90t^2
6.05 = 0.866Vo*t
Vo = 7.0/t
1.45 = 3.54.9t^2
t = √2.05/4.9 = 0.65 sec
Vo = 6.05/(0.866*0.65) = 10.8 m/sec
60° angle
(3.051.6) = Vo*0.866*t4.90t^2
6.05 = 0.5Vo*t
Vo = 12.1/t
1.45 = 12.1*0.8664.9t^2
t = √9.0/4.9 = 1.36 sec
Vo = 6.05/(0.5*1.36) = 8.90 m/sec

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 recalculate
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Ben say: Only 1.6 m above the ground? That's a pathetic jump shot.

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