A. Horizontally Thrown Projectile

1. A rifle was fired horizontally and travels 200.00 meters. The rifle barrel is 1.90 meters from the ground. What speed must the bullet have been travelling at?

Given: Formula: Solution: Answer:

2. A ball rolls with a speed of 2.0 m/s across a level table that is 1.0 meter above the floor. Upon reaching the edge of the table it follows a parabolic path to the floor. How far the floor is the landing spot from the table?

Given: Formula: Solution: Answer:

B. Projectile Launched at an Angle

1. A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 30.0 from the horizontal with a velocity of 40.0 m/s. How far will it travel in the air?

Given: Formula: Solution: Answer:

Here some handouts and sample problems to help you solve these problems.

Horizontally Thrown Projectile

Vix= 2.4 m/s Viy- 0 m/s ax= 0 ay= - 9.8 m/s t= ? t=? Note: In order to find the distance, solve first for the time. Formula dy = v t + ayt /2 Solution: 60 m = (0m/s)t + (- 9.8 m/s?)t?/2 0.60m (-4.9 m/s?)t2 -4.9 m/s -4.9 m/s VO. 122 s2 = Vt2 Answer: t= 0.3493 s Formula: d. = v t + axt?/2 d. = v t ix d.= 2.4m S (0.3493s) Answer: d = 0.83 mAnswer: dmax= 15.94 mViy= 25 m/s Sin 450 = 17.68 m/s Formula: for solving the time of flight Vy = Viy + at Solution: Vy = Viy + a t 0= 17.68 m/s + (-9.8 m/s') t (-9.8 m/s )t 17.68 m/s (-9.8 m/s ) (-9.8 m/s?) Answer: t= 1.80s Formula: for solving the distance d =vt Solution: d = 17.68 (1.80s) Answer: d = 31.82 m Formula: for solving the peak height dmax= (Vy?)-Voy? 2g Solution: dmax= (Vy ? )-Voy? 2g S (17.68 m/s) - (0 m/s)2 2( -9.8 m/s) 312.5824 m /s -19.6m/sHorizontally Thrown Projectile - A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight and the horizontal distance of the projectile. EQUATIONS: HORIZONTAL VERTICAL d = v. t + axt /2 dy = v t + ayt /2 Where: v. = initial horizontal velocity (m/s) v = initial vertical velocity (m/s) ax= 0 dx= distance/displacement Viy= 0 m/s ay= -9.8 m/s2 Sample Problem A pool ball leaves a 0.6 meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location. Given: HORIZONTAL VERTICAL d. = ? d = 0.60 mProjectile Launched an Angle - A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally. Upon reaching the peak, the projectile falls with a motion that is symmetrical to its path upwards to the peak. Predictable unknowns include the time of flight, the horizontal range, and the height of the projectile when it is at its peak. EQUATIONS: HORIZONTAL VERTICAL Vx = Vix + at Vy = Viy + a t d =v.t+ axt /2 Viy= Sin e Vix=Cose dmax= ( Vy') -Voy 2g v = initial horizontal velocity (m/s) v_= initial vertical velocity (m/s) ax = 0 dx= distance/displacement Viy= 0 m/s ay= -9.8 m/s e= angle given dmax= maximum/peak height Sample Problem A football is kicked with an initial velocity of 25m/s at an angle of 450 with the horizontal. Determine the time of flight the horizontal distance and the peak height of the football. Given: HORIZONTAL VERTICAL dx= ? Viy= 21.27 m/s Vi= 25m/s ay= -9.8 m/s Vix=Cos 450= 17.68 Viy= Sin 450= 17.68 m/s m/s Vy= 0 Solve for Vix and Viy: Vi= 25m/s 0 = 450 Vix= 25 m/s Cos 450 = 17.68 m/s