High Jumper Acceleration Calculation

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To calculate the horizontal and vertical acceleration of a 76kg high jumper experiencing a ground reaction force (GRF) of 1750N at a 70-degree angle, one must first resolve the GRF into its x (horizontal) and y (vertical) components using trigonometry. The horizontal component is found by multiplying the GRF by the cosine of the angle, while the vertical component is determined by multiplying the GRF by the sine of the angle. After calculating the net forces in both directions, apply Newton's second law (net force = mass * acceleration) to find the respective accelerations. The jumper's vertical acceleration must also account for the downward force of gravity, which is 9.81 m/s². Understanding these calculations is crucial for determining the jumper's performance during takeoff.
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Hi, I am really stuck on this question, any help would be really really helpful!

During takeoff a 76kg high jumper experiences a GRF of 1750N acting at an angle of 70 degrees to the horizontal. Taking acceleration due to gravity to be 9.81 ms and ignoring air resistance, calculate the HORIZONTAL and VERTICAL accleration experienced by the jumper at this point in time.

I think i need to work out the net force first, but after this I am completely confused. Thanks :frown:
 
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1. calc net force in x- and y composants

2. Net force = mass * acceleration, Newtons famous law.
 
i understand the equation and got the net force to be 745.6 ? but i dnt understand how to fo it for x (horizontal) and Y (vertical). I know this probs sounds daft but physics really isn't my strong point! Thanku
 
in what direction?...

and what is GRF ?
 
and yeah, use trigonometry.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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