# webct1 - kid-ball 1 A kid throws a ball in the air with an...

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kid-ball 1. A kid throws a ball in the air with an initial velocity of 6 m/s directed upwards. The ball leaves the hand of the child at 1.3 m from the ground. Take the upward direction to be positive. What was the speed of the ball at the highest point of the trajectory? The speed of the ball at the highest point of the trajectory is 0 m/s. It is instantaneously at rest as it goes from a “positive” velocity (vector pointing up) to a “negative” velocity (vector pointing down). 2. What is the maximum height from the ground that the ball will reach? Let us take y = 0 at ground level and the + y axis pointing up. Then, 0 0, 2 1.3 m 6 m/s 9.8 m/s y y y v a = = =− The condition for maximum height is v y = 0, so 22 0t o p 0 2 0 top 0 2 2( ) 0( 6 m / s ) 1.3 m 3.1 m 2 2(9.8 m/s ) y y vv a y y yy a −= =+ = + = 3. What is the magnitude of the acceleration of the ball at the peak of the trajectory? The acceleration due to gravity is constant, so 9.8 m/s 2 . 4. What is the speed of the ball right before hitting the ground? The condition for “hitting the ground” is y = 0 (with our choice of origin at ground level).

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22 ground 0 ground 0 2 ground 0 0 2( ) 2 For the velocity, we should use the sign, because when the ball hits the ground, the y-component of velocity is negative (because we chose +y pointing y y vv a yy va y v −= =± − + ground up). 2( 9.8 m/s )(1.3 m) (6 m/s) 7.8 m/s So the speed is 7.8 m/s. v =− − − + =−
xt 1. A car travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x ( t ) = bt 3 - c t 2 , where b = 0.14 m/s 3 and c = 2.6 m/s 2 . How long after the stop does the car stop again? The first derivative of the displacement equation will give the velocity equation. 2 () 3 2 dx vt b t c t dt == The car stops again when the velocity becomes zero: 2 32 0 2 Two solutions: 0 and 3 bt ct c tt b −= t = 0 corresponds to the initial point (when the car is at the stop sign), so we want the other solution. 2. Find the instantaneous velocity at t = 3 s. 2 3 2 dx b t c t dt At t = 3s, we have ( b = 0.14m/s 3 and c = 2.6 m/s 2 ) 2 (3) 3(0.14 m/s )(3 s) 2(2.6 m/s )(3 s) 11.8 m/s v =− = 3. Find the instantaneous acceleration at t = 3 s. 2 2 6 2 dx at b t c dt At t = 3s, we have ( b = 0.14m/s 3 and c = 2.6 m/s 2 ) 2 (3) 6(0.14 m/s )(3 s) 2(2.6 m/s ) 2.68 m/s a =

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4. Find the average velocity between t = 2 s and t = 4 s. final initial average final initial xx x v tt t == ∆− We need the position of the car at t = 2 s and t = 4 s (use the x ( t ) expression above) average (2 s) 9.28 m (4 s) 32.6 m 32.6 ( 9.28) 11.7 m/s 42 x x v =− −− 5. Find the average acceleration between t = 2 s and t = 4 s.
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## This homework help was uploaded on 03/26/2008 for the course PHYS 221 taught by Professor Herrera-siklody during the Spring '08 term at Iowa State.

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webct1 - kid-ball 1 A kid throws a ball in the air with an...

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