Given vi = 10 m/s If the projectile is intended to hit a target + 12.8 m downrange (delta x), 1) how tall should the launcher be?  give launch degree =0 (launch horizontally)   should I use horizontal motion of a projectile equation? 2) is there any vi x and vi y value?

An arrow is launched from ground level, with an initial speed of 27.2 m/s at an angle of 31.4° above the horizontal. Neglect air resistance, and take upward as the positive direction. a) How much time is needed for the arrow to reach its maximum height, b) what is this maximum heightc) how long does it take to land on the ground againd) how far does it travel right before it lands.

1. The distance between home plate and the pitcher's mound in baseball is 60.5 ft. In MLB, baseballs are. pitched at about 90mph. Assume the ball is thrown perfectly horizontally, at 0°. How far in feet will the ball drop during its transit to home plate? Use g = 32 ft/s². (Note: this is why pitcher's mounds are raised.) 2. A pebble is tossed into the air at an angle of 62° above the horizontal, with an initial speed of 4.8 m/s. The pebble is 1.6m above the ground when it is released by the thrower's hand. a) How long will the pebble remain airborne? b) What maximum height above the ground will the pebble achieve?

Chinook salmon can jump out of water with a speed of 7.50 m/s. How far horizontally d can a Chinook salmon travel through me air if it leaves the water with an initial angle of 0 = 26.0° with respect to the horizontal? (Neglect any effects due to ir resistance.) m d = Question Source: Freedman College Physics Se

A projectile is fired with initial speed 150 m/s and an angle of elevation 45° from a position 10 m above ground level. Where does the projectile hit the ground and with what speed? Solution If we place the origin at ground level, then the initial position of the projectile is  (0, 10)  and so we need to adjust the parametric equations of the trajectory by adding 10 to the expression for y. With  v0 = 150 m/s,  ? = 45°, and g = 9.8 m/s2, we have x  =  150 cos    ? 4    t =          y  =  10 + 150 sin    ? 4    t −  1 2  (9.8)t2 =          . Impact occurs when y = 0, that is  4.9t2 − 75   2  t − 10 = 0.  Solving this quadratic equation (and using only the positive value of t), we get the following. (Round your answer to two decimal places.) t =  75   2  +    11,250 + 196 9.8  ≈   Then  x ≈ 75   2 (21.74) ≈    (rounded to the nearest whole number), so the projectile hits the ground about…

A target lies a distance R = 1100 m from a cannon across level ground, and a distance d = 780 m behind a wall h = 400 m tall.  If the cannon shoots projectiles with speed 160 m/s, at what angle should the cannon fire so as to hit the target over the wall?

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Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a spitting cobra rears up to a height of 0.520 m above the ground and launches venom at 4.00 m/s, directed 41.0° above the horizon. Neglecting air resistance, find the horizontal distance (in m) traveled by the venom before it hits the ground. HINT m Need Help? Read it Watch It

The initial speed of a tennis ball is 57.5 m/s and the launch angle is ?i = 16°. Neglect air resistance.Question 1) What is the maximum height, h, of the tennis ball? in mQuestione 2) What is the range, R, of the tennis ball? in m info given: what angle results in the greatest height=90 For this 90° angle, the horizontal component of velocity is zero, so the initial velocity is completely in the vertical direction, resulting in greatest height Kato tries substituting ty,max for t, 0 for yi, and h for yf, and gets h =  vi2 sin2 ?i 2g   When ?i = 90°, sin2 ?i is maximum, so h is maximum what angle results in the maximum horizontal range=45°  I substituted 2vi sin ?i g into the expression for the horizontal component of velocity, (vi cos ?i)t, and got  R =  2vi2 sin ?i cos ?i g when ?i = 45°, 2?i = 90°, and sin 90° = 1, which is its maximum value.

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