A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 13.35 m/s at an angle of 25.1 degrees below the horizontal. It strikes the ground 3.53 s later. Calculate the height from which the ball was thrown. Your Answer: units Answer
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- When a ball with a mass of 3 kg reaches 9 meters above the ground after being thrown obliquely from the ground to the air, the velocity vector is calculated with v = 7i + 6j. Accordingly, what is the maximum height the ball can reach?A projectile is launched on the Earth with a certain initial velocity and moves without air resistance. Another projectile is launched with the same initial velocity on the Moon, where the acceleration due to gravity is one-sixth as large. How does the range of the projectile on the Moon compare with that of the projectile on the Earth? (a) It is one-sixth as large. (b) It is the same. (c) It is 6 times larger. (d) It is 6 times larger. (e) It is 36 times larger.A ball thrown horizontally with an initial velocity of = 23.0 m/s travels a horizontal distance of d = 58.0 mm before hitting the ground. From what height h was the ball thrown? using the equation provided solve for h which is the same as Yinitial Yfinal = 0 Yinitial = ? Initial velocity of y = 9.8 m/s gravity = 9.8 m/s Change in time = 35 s
- A speedboat increases its speed uniformly from vi= 20m/s to vf= 30m/s in a distance of 2.00x102m. Draw a coordinate system for this situation and label the relevant quantities, including vectors.A stone is thrown outward, at an angle of 30° with the horizontal into the river from a cliff, that is 75 meters above the water level at a velocity of 54 km/hr. How long it will take the stone to hit the surface of the river? a. 0.06159 minb. 0.059157minc. 0.01596 mind. 0.07915 minA projectile is projected from the origin with a velocity of 30.0 m/s at an angle of 40.0 degrees above the horizontal. What is the time it takes the projectile to reach maximum height? Multiple Choice 2.23 s 1.02 s 1.97 s 1.54 s 2.66 s
- A horizontal projectile is shot with an initial velocity of 17.9 m/s from a height of 2.59 m above the ground. Neglecting air-resistance, what is the magnitude of the final velocity (m/s)The drawing shows two planes each dropping an empty fuel tank. At the moment of release each plane has the same speed of 166 m/s, and each tank is at the same height of 2.68 km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of 15.0° above the horizontal and the other is flying at an angle of 15.0° below the horizontal. Find the (a) magnitude and (b) direction of the velocity with which the fuel tank hits the ground if it is from plane A. Find the (c) magnitude and (d) direction of the velocity with which the fuel tank hits the ground if it is from plane B. In each part, give the direction as a positive angle with respect to the horizontal.A brick is thrown upward from the top of a building at an angle of 30° to the horizontal and with an initial speed of 12 m/s. If the brick is in flight for 2.6 s, how tall is the building? step-by-step
- At t=0, a ball is thrown with an initial velocity of magnitude 15ms-1 at 75° to the horizontal (upwards). The magnitudes of the vertical and horizontal components of its initial velocity are 14.49 m/s and 3.88 m/s. The magnitude of the vertical component of velocity of the ball at t=0.5s is 9.585 m/s. The magnitude of the vertical component of velocity of the ball at t=2.0s is 5.13 m/s. Which way is the ball moving at this time? What is the magnitude of the horizontal component of velocity at t=2.0s?In a sport feast, a soccer ball is kicked from ground level over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal component of 30 m/s. Using g = 10 m/s^2, what is the ranged of the soccer ball? A. 40 m. B. 60 m. C. 80 m. D. 120 m. E. 180 m.A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 146 feet per second from an initial height of 89 feet off the ground, then the height of the projectile, h, in feet, t seconds after it's shot is given by the equation: h=−16t2+146t+89h=-16t2+146t+89 Find the two points in time when the object is 126 feet above the ground (to one decimal places). Answer: The object is 126 feet off the ground at the following times: (Enter your two answers separated by a comma. Round to ONE decimal places.)