Projectile fired at an angle from 1m

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celcon
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a projectile is fired upward at an angle, with a speed of 15.0 m/s and from a height of 1.00m. A) if its speed to the top of its arc is 7.5 m/s, what height does it reach? B) what is the speed with which it hits the ground?


I figured the part A to be 9.6m but I am lost on how to figure the speed at which it hits the ground.

Thanks in advance
 
Use kinematics to find the final vertical velocity of the projectile (Vf^2 = Vi^2 + 2gy) at impact. Use this result, and the horizontal component of the velocity = 7.5 m/s, in the Pythagorean Theorem to determine the resultant final velocity.
 
Initial velocity Vi = 15 m/s. At the top of the arc the velocity is horizontal and is equal = Vcos(theta) = 7.5 m/s. From this find the angle of projection and vertical component of the Vi. When it reaches the top of arc Vf = 0. Now using kinematics find h. Don't forget add the initial height.
In part B, the disoplace of the projectile is - 1m. So find the final velocity when it reaches the ground. Its direction is vertical. Hence find the resultant of this velocity and horizontal component to get the final result.