If Roger hit the ball at 45deg, 1ft from t he ground, and it traveled...

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CrzySxyQwel

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...90ft, find the initial speed of the bal? If Roger hit the ball at 45deg, 1ft from the ground, and it traveled 90ft, find
the initial speed of the ball in m/s.
 
The initial velocity of an object is speed of the object before the application of the force.
As we know that force is defined as

f = m×a

where,
m = mass of the object; and
a = acceleration of the object.

So, the final velocity is defined as:

v(f) = v(i) + a×t

where,
v(f) = final velocity of the object;
v(i) = initial velocity of the object; and
a = acceleration of the object in the given time t.
 
From x(T) = Ux T = 90 ft. we have T = 90/Ux as the total flight time.

Then y(T) = h + Uy T - 1/2 g T^2 = h + Uy (90/Ux) - 1/2 32.2 (90/Ux)^2 = 0 = 1 + tan(45) 90 - 16.1 * 90^2/Ux^2 is the height at any time t = T.

So 1 + tan(45) 90 = 16.1 * 90^2/Ux^2 and Ux^2 (1 + tan(45) 90) = 16.1*90^2

And then Ux = sqrt(16.1*90^2/(1 + tan(radians(45))*90)) = 37.85600247 = 37.86 ft/s.

And finally U = Ux/cos(45) = 37.86/cos(radians(45)) = 53.54212547 = 53.5 ft/s. ANS.

NOTE: Disregard radians, that's for Excel to work.
 
it is a projectile question solve it through Newton's equation

let v be the initial velocity

In horizontal direction
s = ut + 1/2 at^2
here a = 0
90 = v * cos45 * t + 0
t = 90*root 2 /v

In vertical direction
s = ut + 1/2 at^2
1 = v*sin45 * 90*root2/v - 1/2 * 9.8 * 90*90*2 / v^2
v = root( 4.9 * 8100 / 89 )
v = 21.12 ft/s
v = 7 m/s

* are you sure about the values
 
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