Give 5 situation and formula about projectile motion.
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Give 5 situation and formula about projectile motion.1.2.3.4.5.
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Answer:
Sure, here are five situations and formulas related to projectile motion:
1. A ball is thrown straight up into the air with an initial velocity of 20 m/s. What is the maximum height it will reach before falling back to the ground?
Formula: h = (v0^2 sin^2θ)/2g, where h is the maximum height, v0 is the initial velocity, θ is the angle of projection (in this case, 90 degrees), and g is the acceleration due to gravity.
2. A football is kicked at an angle of 30 degrees above the horizontal with an initial velocity of 25 m/s. How far will it travel before hitting the ground?
Formula: R = (v0^2 sin 2θ)/g, where R is the horizontal range, v0 is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.
3. A cannonball is fired from a cannon at an angle of 45 degrees above the horizontal with an initial velocity of 100 m/s. What is its velocity at the highest point of its trajectory?
Formula: vy = v0 sin θ - gt, where vy is the vertical component of the velocity at any time t, v0 is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity. At the highest point of the trajectory, vy = 0, so we can solve for t and then substitute it into the formula vx = v0 cos θ to find the horizontal component of the velocity.
4. A baseball is thrown horizontally off a cliff with an initial velocity of 30 m/s. How long will it take to hit the ground below?
Formula: t = sqrt(2h/g), where t is the time of flight, h is the height of the cliff, and g is the acceleration due to gravity. Since the ball is thrown horizontally, its initial vertical velocity is zero.
5. A rocket is launched vertically upwards with an initial velocity of 80 m/s. What is its velocity after 5 seconds?
Formula: v = v0 - gt, where v is the velocity at any time t, v0 is the initial velocity, and g is the acceleration due to gravity. Since the rocket is launched vertically upwards, its initial horizontal velocity is zero.
Verified answer
Answer:
1. A ball is kicked from the ground with an initial velocity of 25 m/s at an angle of 40 degrees above the horizontal. What is its maximum height and how far does it travel horizontally before hitting the ground?
max height = ((25)^2 sin^2(40))/2(9.81) = 31.8 m
distance = (25^2 sin(80))/9.81 = 108 m
2. A cannonball is fired from a height of 50m above the ground with an initial velocity of 60 m/s at an angle of 30 degrees above the horizontal. How long does it take to hit the ground and what is its velocity when it hits?
time of flight = (2v*sin(30))/g = 7.4 s
velocity = sqrt(vx^2 + vy^2 + vz^2) = sqrt((60*cos(30))^2 + (60*sin(30) - 9.81*7.4))^2 + 0^2) = 186 m/s
3. A rocket is launched from a platform and has an acceleration of 20 m/s^2. If it is launched at an angle of 60 degrees above the horizontal and reaches a height of 500 meters, what was its initial velocity and how far did it travel horizontally?
initial velocity = sqrt(h*g/2*(1-cos(60))) = 200 m/s
distance = (v^2*sin(120))/g = 6667 m
4. A quarterback throws a football from the 30-yard line to his receiver on the opposite 20-yard line. If the ball is thrown at a velocity of 25 m/s at an angle of 45 degrees above the horizontal, what is the time of flight and how high does it go?
time of flight = (2v*sin(45))/g = 3.2 s
max height = (v^2*sin^2(45))/2g = 15.6 m
5. A baseball is hit from home plate with an initial velocity of 40 m/s at an angle of 35 degrees above the horizontal. If the field is level and the ball lands 120 meters away, what was its initial height and what was its time of flight?
initial height = (v^2*sin^2(35))/(2g) = 19.4 m
time of flight = (2v*sin(35))/g = 4.4 s
Explanation:
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