Its initial speed cannot be 20 m, as stated in the question. Secondly, if the initial speed is correctly given, then there is no need to calculate it!
Yes.
We have no idea how big the rock is, and no way to figure it out. But we can calculate that it reaches 11.48 meters above the ground before it starts falling.
Speed(74) = 72.2Speed(50) = 44.7
If you know the initial speed (u), acceleration (a) and time (t), then the final speed, v = u + at.
True
No. The one with higher initial speed will hit the ground first if they are both thrown straight down.
Yes.
A projectile that is thrown with an initial velocity,that has a horizontal component of 4 m/s, its horizontal speed after 3s will still be 4m/s.
A ball is thrown vertically upward with an initial speed of 20m/s. Two second later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24m/s. At what height above the release point will the ball and stone pass each other?
6.261 m/s
distance divided by speed is equal to timetraveled.
We have no idea how big the rock is, and no way to figure it out. But we can calculate that it reaches 11.48 meters above the ground before it starts falling.
If it is thrown at an angle, at the top of its path, its vertical velocity will be zero, however its horizontal velocity will be the same as its initial horizontal velocity minus whatever loss in speed as a result of air friction at that point. We won't know what that is without more information.
Speed(74) = 72.2Speed(50) = 44.7
If you know the initial speed (u), acceleration (a) and time (t), then the final speed, v = u + at.
True
Speed(74) = 72.2Speed(50) = 44.7