Calculating Projectile Motion and Speed: Vertical and Horizontal Components

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To calculate the projectile motion needed to clear an 80m obstacle, the vertical speed and time to reach maximum height must be determined first. The vertical component of the initial velocity can be calculated using the equation Vvertical = vsinΘ, where Θ is the launch angle. At maximum height, the vertical velocity becomes zero, indicating the point where the projectile stops rising. The horizontal distance of 100m allows for the calculation of the horizontal speed needed for the projectile to travel that distance. Overall, understanding the relationship between vertical and horizontal components is crucial for solving the problem effectively.
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A projectile is to be fired so that it just clears an obstacle 80m high at its maximum height.
http://img167.imageshack.us/img167/5729/physicszh7.jpg
a) by considering the vertical motion only, calculate the verticle speedand the time taken to reach its maximum height
b) if the distance between A to B is 100m, calculate the horizontal speed with which the ball was thrown.
c) Hence find v and Θ.


totally stuck at this, any help much appricated.
 
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Have you tried anything? What equations might be used? What can you say about the vertical component of the velocity at maximum height?
 
the farthest i got, was putting in the right angle, and as for equations that could be used, i think it may have something to do with - Vvertical= vsinΘ. not entirely sure though.
 
That would be the initial vertical velocity (is that a negative sign?). What is the vertical velocity at the instant of maximum height?
 

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