Answer:

lTo solve these problems, we will use the equations of motion for a projectile in free fall, neglecting air resistance.

1.To find the height of the tower, we can use the equation for the horizontal range of a projectile: range = horizontal velocity x time Since the ball was thrown horizontally, its initial vertical velocity is zero, so it will take the same amount of time to hit the ground as an object dropped from the same height. Therefore, we can use the equation for the vertical displacement of a falling object: vertical displacement = (1/2) x acceleration x time^2 where acceleration due to gravity is -9.8 m/s^2 (negative because it is downward). We can set these two equations equal to each other and solve for the height of the tower: range = horizontal velocity x time 40 m = 8 m/s x time time = 5 s vertical displacement = (1/2) x acceleration x time^2 vertical displacement = (1/2) x (-9.8 m/s^2) x (5 s)^2 vertical displacement = 122.5 m Therefore, the height of the tower is 122.5 m.

2.To find the time the Frisbee stays in the air, we can use the equation for the horizontal range of a projectile again: range = horizontal velocity x time Since the Frisbee was thrown horizontally, its initial vertical velocity is zero, so the time it stays in the air will be determined by its vertical motion, which is governed by the same equation as before: vertical displacement = (1/2) x acceleration x time^2 where acceleration due to gravity is -9.8 m/s^2 (negative because it is downward). We know that the Frisbee moves a horizontal distance of 15 m, so we can use this to solve for the time it stays in the air: range = horizontal velocity x time 15 m = 12 m/s x time time = 1.25 s vertical displacement = (1/2) x acceleration x time^2 vertical displacement = (1/2) x (-9.8 m/s^2) x (1.25 s)^2 vertical displacement = 7.69 m Therefore, the Frisbee stays in the air for 1.25 s.

3.To find the distance the jackstone ball travels before it hits the ground, we can use the equation for the horizontal range of a projectile again: range = horizontal velocity x time Since the ball was thrown horizontally, its initial vertical velocity is zero, so the time it stays in the air will be determined by its vertical motion, which is governed by the same equation as before: vertical displacement = (1/2) x acceleration x time^2 where acceleration due to gravity is -9.8 m/s^2 (negative because it is downward). We know that the window is 3 m above the ground, so we can subtract this from the vertical displacement to get the height the ball falls from: vertical displacement = (1/2) x acceleration x time^2 vertical displacement = (1/2) x (-9.8 m/s^2) x (time)^2 vertical displacement = 0 - 3 m time = 0.62 s Now we can use the equation for range to find the distance the ball travels before it hits the ground: range = horizontal velocity x time range = 14.42 m/s x 0.62 s range = 8.93 m Therefore, the ball travels 8.93 m before it hits the ground.

4.To find the horizontal initial velocity of the projectile, we can use the equation for the horizontal range of a projectile: range = horizontal velocity x time Since the projectile was fired horizontally, its initial vertical velocity is zero, so the time it takes to reach the ground is determined by its vertical motion, which is governed by the equation for the vertical displacement of a falling object: vertical displacement = (1/2) x acceleration x time^2 where acceleration due to gravity is -9.8 m/s^2 (negative because it is downward). We know that the projectile falls a distance of 19.62 m, so we can use this to solve for the time it takes to reach the ground: vertical displacement = (1/2) x acceleration x time^2 19.62 m = (1/2) x (-9.8 m/s^2) x time^2 time = 2 s Now we can use the equation for range to find the horizontal initial velocity of the projectile: range = horizontal velocity x time 20 m = horizontal velocity x 2 s horizontal velocity = 10 m/s Therefore, the horizontal initial velocity of the projectile was 10 m/s.