Please provide a step by step solution. Thank you so much for your kind support.

Question

Please provide a step by step solution. Thank you so much for your kind support.

You throw a ball towards a wall at speed 24 m/s and at an angle 40.0 degree above The horizontal. The wall is 20.8 m from The release point of The ball. How long does The ball take to reach The wall? s How far above The release point does The ball hit The wall? what are The horizontal and vertical components of its velocity as it hits The wall? horizontal m/s vertical m/s when it hits, has it passed The highest point on its trajectory? yes no not enough information to decide

Explanation / Answer

vi=24 m/s
=40°
x=20.8m

0) Use this equation because you have x component and not y.
You need time or part c so you solve it like this
x=(vi*cos)t
20.8=(24cos40°)t
t=1.13166 s

a) This is a trajectory and you need y so you use this equation.
y=(tan)x-(gx^2/(2(vi*cos)^2)
y=tan40°*20.8m-((-9.8*20.8m)^2)/(2(24*cos40°) ^2...
y=17.45327m-61.463m
y=-44.0m= -44.0m

b) Break it down into it's components.
vx=cos*vi
vx=cos40°*24
vx=18.385m/s = 18.385m/s

c) vy= sin 40°*24-.5(-9.8)(1.13166s)
vy=20.973m/s

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