A circus performer launches himself from a springboard with an initial velocity

Question

A circus performer launches himself from a springboard with an initial velocity of 21 m/s at an angle of 75 degrees toward a platform hanging 20m above the ground, a horizontal distance of 15 m away from the springboard. Does he land on the platform or fall back down to the ground? Make and model with graphs and motion maps, including the maximum height he reaches, the time of flight and his final velocity (direction and magnitude) just befor elanding. show all work and explain your reasoning. take g= 9.8 m/s^2

Explanation / Answer

here,

vx = 21cos75 = 21*0.25 = 5.25 m/s
Vy = 21sin75 = 21*0.96 = 20.16 m/s

equation of trajectory

y = xtan - (g*x^2) / (2*v^2*cos^2)

x is the horizontal component,
y is the vertical component,
g= gravity value,
v= initial velocity,
= angle of inclination of the initial velocity from horizontal axis,

y = 15*tan75 - (9.8*15^2) / (2*15^2*cos^2 75)

y = -17.6 m

as this is less than 20m so he will not able to reach platform

as

time = distane / speed = 15 / Vsin75
t = 15 / 21*sin75
t = 0.73s

time for flight is 0.73 s

Mamimum Height H = V^2*(sin^2 75) / 2g

H = ( 21^2 * (sin^2 75) ) /2 *9.8 = 20.99 m

maximum height reach by performer is 20.99m

final velocity just before reaching the ground

V^2 - u^2 = 2* g * S
(Vsin75)^2 = 2 * 9.8 * 15

V = 17.75m/s and -17.75m/s

final velocity just before reaching the ground is 17.75m/s and -17.75m/s

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