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Vertical projectile motion

Vertical projectile motion

Assessment

Presentation

Physics

10th - 11th Grade

Medium

Created by

Stacy King

Used 3+ times

FREE Resource

4 Slides • 3 Questions

1

Vertical projectile motion

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2

change in time for vertical projectile motion

  •  y=vitsinθ+12gt2y=v_it\sin\theta+\frac{1}{2}gt^2  

  • so

  •  Δt=2visinθ(g)\Delta t=\frac{2v_i\sin\theta}{\left(-g\right)}  

3

Fill in the Blank

A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight. hundreths

4

horizontal (x) displacement

  •  x=vitcosθx=v_{i_{ }}t\cos\theta  

  • so

  •  Δx=vi2sin(2θ)(g)\Delta x=\frac{v_i^2\sin\left(2\theta\right)}{\left(-g\right)}  

5

Fill in the Blank

A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the horizontal displacement. hundreths

6

maximum height

  •  vy=visinθ+gtv_y=v_i\sin\theta+gt  

  •  vy=0v_y=0  at max height

  •  tvy=Δt2t_{vy}=\frac{\Delta t}{2}  

  •  tvy=visinθ(g)t_{vy}=\frac{v_i\sin\theta}{\left(-g\right)}  sub into DE

  •  hmax=vi2sin2θ(g)+12g(vi2sin2θ)g2h_{\max}=\frac{v_i^2\sin^2\theta}{\left(-g\right)}+\frac{\frac{1}{2}g\left(v_i^2\sin^2\theta\right)}{g^2}  

  •  hmax=vi2sin2θ2(g)h_{\max}=\frac{v_i^2\sin^2\theta}{2\left(-g\right)}  

7

Fill in the Blank

A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the max height. tenths

Vertical projectile motion

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