Projectile Motion Help!!

Question

I completed a lab where in a group each member took a turn jumping and the length of the jump as well as the time in the air during the jump. I'm stuck on my final question and am not sure how to calculate this. It states: using both these measurements, calculate the magnitude and direction of the take-off velocity.

 

Jump distance: 1.83 m

Time in air: 0.55 s

resultant magnitude ?? m/s

resultant angle: ?? º

Answer 1

Lets do it from basics.

Suppose your take off velocity is v m/s with a take off angle w with respect to ground

 

As per convention assume x axis be along the ground and y axis upwards, and you atrt from (0,0)

 

Upward component of velocity is vy = vsina

Forward component of velocity vx=vcosa

 

Time in air = .55s, So time taken for vy to become 0 duet to gravity is .55/2 = .275 s

 

vy = vsina =  g * .275 = 9.8 *.275 =  2.695

 

Jump distance = 1.83m ie distance tarveled at vx speed in .55s.

so 1.83 = .55 * vcosa or vcosa = 1.83/.55 = 3.327

v^2 = vx^2 + vy^2 =3.327 ^2 + 2.695 ^2 = 18.337

v = sqrt(18.337) = 4.28

a = sininv(2.695/4.28) = 39 degree

 

Take off velocity was 4.28 m/s at 39degree with respect to ground.