Joe stands on a bridge kicking stones into the water below. If he kicks a stone with a horizontal velocity of 2.10 m/s, and it lands in the water a horizontal distance of 7.50 m from where Joe is standing, what is the height of the bridge? Enter m as unit.
I don't understand... I think I did everything right, followed the formulas correctly I think... the answer I got was 5.3 m, or something along those lines. Please someone show me how to do this? I promise 10 points for your time and effort!
I don't understand... I think I did everything right, followed the formulas correctly I think... the answer I got was 5.3 m, or something along those lines. Please someone show me how to do this? I promise 10 points for your time and effort!
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If we assume that the horizontal speed of the stone is constant, we can find the time taken for the stone to land in the water:
speed = distance / time
2.1 = 7.5 / t
t = 7.5 / 2.1
Then assuming that the stone had no initial vertical speed, we can say that vertically, the stone fell from rest over 7.5 / 2.1 seconds.
Using S = ut + 0.5at^2
S = 62.5m
speed = distance / time
2.1 = 7.5 / t
t = 7.5 / 2.1
Then assuming that the stone had no initial vertical speed, we can say that vertically, the stone fell from rest over 7.5 / 2.1 seconds.
Using S = ut + 0.5at^2
S = 62.5m
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First figure out how long it takes the stone to fall. Since it continuously travels at 2.10 m/s horizontally until it lands...
distance(horizontal) = velocity(horizontal) x time (constant velocity)
d(x) = v(x) * t
solve for t...
t = d(x)/v(x)
then write the equation for the distance it fell due to acceleration
distance (vertical) = 0.5 * acceleration * square (time)
d(y) = 0.5 * at^2
substitute for t from the first equation
d(y) = 0.5 * a * [d(x)/v(x)]^2
= 0.5 * (9.81 m/s^2) * (7.50 m / 2.10 m)^2
= 62.6 m
distance(horizontal) = velocity(horizontal) x time (constant velocity)
d(x) = v(x) * t
solve for t...
t = d(x)/v(x)
then write the equation for the distance it fell due to acceleration
distance (vertical) = 0.5 * acceleration * square (time)
d(y) = 0.5 * at^2
substitute for t from the first equation
d(y) = 0.5 * a * [d(x)/v(x)]^2
= 0.5 * (9.81 m/s^2) * (7.50 m / 2.10 m)^2
= 62.6 m
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Applying law's of motion in the horizontal direction,
s = ut + (1/2)a×t^2
7.5 = 2.1t + (1/2)×0×t^2 ,since there is no acceleration in the horizontal direction
Gives,
t = 3.57 sec
Applying law's of motion in the vertical direction,
s = 0×3.57 + (1/2)×9.8×3.57^2
=62.5m (answer)
For all physics assignment help and detailed solution visit www.tutornik.com
s = ut + (1/2)a×t^2
7.5 = 2.1t + (1/2)×0×t^2 ,since there is no acceleration in the horizontal direction
Gives,
t = 3.57 sec
Applying law's of motion in the vertical direction,
s = 0×3.57 + (1/2)×9.8×3.57^2
=62.5m (answer)
For all physics assignment help and detailed solution visit www.tutornik.com
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for horizontal motion
a(x) = 0 ,u(x) = 2.10m/s , x=7.50
x=ut+1/2at^2
7.50=2.10*t
t = 3.57s
for vertical motion
a(y) = g , u(y) = 0 , t=3.57s
y = ut+1/2at^2
y = 1/2*9.8*(3.57)^2
= 62.5m
so bridge is 62.5m high
a(x) = 0 ,u(x) = 2.10m/s , x=7.50
x=ut+1/2at^2
7.50=2.10*t
t = 3.57s
for vertical motion
a(y) = g , u(y) = 0 , t=3.57s
y = ut+1/2at^2
y = 1/2*9.8*(3.57)^2
= 62.5m
so bridge is 62.5m high
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God i hate math