rectilinear motion

Question

A ball is thrown horizontally towards a wall.Given that the ball needs 0.5s to reach the wall and the distance between the original position of the ball and the wall is 20m.
Find the speed of the ball when it reach the ball.
Assume that the speed of the ball is constant when it travels.

I use
s=1/2(u+v)t
20=1/2(v)(0.5)
v=80 ms-1

when I use
v=s/t
v=20/0.5
v=40 ms-1

Why different answers are obtained in my calculation?

Answer 1

唔~~我覺得個題目本身有d 問題. 因為一開始你 "Assume that the speed of the ball is constant when it travels", 咁即係個波全程都係同一個speed, 咁係呢一個assumption 上面, 咁我地就應該用 method 2.

但係呢一個答法都重要睇吓係幾年班啦, 因為如果只係小學level, 咁當然可以用最簡單的公式 s=vt, 因為果時計的數係無acceleration 的.

但係如果係 form 4 或以上, 咁就要睇埋地心吸力的影響啦. 因為當我地睇一個物體的speed / velocity, 我地都要睇2 個perpendicula directions (今次來講即係向橫同向下), 而向下係有地心吸力, 所以個向下的speed 係要用 v= u+ at 來計的. (u=0, a = 10, t = 0.5), 咁我地就會計到 v = 5ms-1. 而且個value 重要係一路咁變緊的, 所以根本無可能係constant speed.
即使我當佢係話個horizontal component 係constant, 咁個final speed 都重要用pyth thm. 去計番出來. 即係 v = sqrt(5^2+40^2) = 40.31ms-1

希望我個答案可以幫到你啦

Answer 2

In the first method, you calculated the final velocity of the ball and that is the velocity of the ball when it arrives the wall. So it is the correct answer.

In the second method, you calculated the average velocity of the ball and which is the velocity of the ball in the mid-way(the mid-point between the starting point and the wall).