An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. a) What is the maximum height reached by the object? b) What is the total flight time (between launch and touching the ground) of the object? c) What is the horizontal range (maximum x above ground) of the object?

An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. a) What is the maximum height reached by the object? b) What is the total flight time (between launch and touching the ground) of the object? c) What is the horizontal range (maximum x above ground) of the object?

Similar questions

From a height of 2 m with respect to the ground, a ball is thrown with a speed V0=30m/s at an angle of θ0=300 with respect to the horizontal. Determine. a) The maximum height that the ball reaches with respect to the ground. b) The time it takes to hit the ground. c) The distance it travels horizontally.
A projectile is launched at ground level with an intial vertical velocity of 25.0m/s and a horizontal velocity of 43.3m/s. It strikes a target above the ground 3.00 seconds later. a) How high is the target above the gound? b) How far did the projectile travel before hitting the target? c) What is the horizontal component of the velocity of the projectile right as it hits the target? d) What is the vertical component of the velocity of the projectile right as it hits the target?
Problem: For a projectile lunched with an initial velocity of v0 at an angle of θ (between 0 and 90o) , a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of θ gives the highest maximum height? Solution The components of v0 are expressed as follows: vinitial-x = v0cos(θ) vinitial-y = v0sin(θ)
You throw a ball up at some angle above the horizontal so that its velocity is (v0x, v0y). You then run 22.0 m forward on level ground and catch the ball 2.90 s later at the same altitude it was released from. a) How high above the release point did the ball reach? b) What was the horizontal component of the ball’s initial velocity, v0x? c) What was the vertical component of the ball’s initial velocity, v0y?
A ball is launched from a 10.0 m building with an initial speed of 15.0 ?/?. If the ball’s speed when it reaches its maximum height is 1/3 its speed when it is at half its maximum height, determine ball’s launch angle and the ball’s angle at impact. This is all the information i was given!
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.
Problem 4 An object is launched at a velocity of 25 m/s in a direction making an angle of 20° upward with the horizontal. What is the maximum height reached by the object?
Find the magnitude of the object's velocity just before it reaches the ground.
For a projectile lunched with an initial velocity of v0 at an angle of θ (between 0 and 90o) , a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of θ gives the highest maximum height? Solution The components of v0 are expressed as follows: vinitial-x = v0cos(θ) vinitial-y = v0sin(θ) a) Let us first find the time it takes for the projectile to reach the maximum height. Using: vfinal-y = vinitial-y + ayt since the y-axis velocity of the projectile at the maximum height is vfinal-y = Then, = vinitial-y + ayt Substituting the expression of vinitial-y and ay = -g, results to the following: = - t Thus, the time to reach the maximum height is tmax-height = / We will use this time to the equation yfinal - yinitial = vinitial-yt + (1/2)ayt2 if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = vinitial-yt + (1/2)ayt2 substituting, the vinitial-y expression above, results to…
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?
A ball weighing 0.45 kg in a 0-g environment has an initial speed of 30 m/s at an angle 0 = 35°. It hits the ground for a brief second At =0.07s and bounces away with the same speed at the same angle as shown but now it is moving upward. a) Label the initial and final velocity vectors and find their components in the x hat and y hat directions. b) Find the acceleration vector. c) Find the normal force on the ball and draw it in the diagram. Lo ܘܢ
A projectile is fired from point A at the edge of a cliff, whose initial velocity components are v0x = 60.0m / s and v0y = 175m / s. The projectile rises and then falls into the sea at point P. The projectile flight time is 40.0s and it does not experience appreciable air resistance in flight. a) Make a diagram of the problem, b) Determine how far it falls, c) what is the height of the cliff? and d) at what angle was it shot?