Find the horizontal and vertical
displacement after 1,2,3,4,5 seconds,
then the path of the motion of ball.
If the ball reaches the ground in 4
seconds, find the height of the tower.
Answers
Dear Friend
For horizontal distance, gravity plays no role.
Horizontal Distance=vtSo, in 1,2,3,4,5 sec the distance will be40m, 80m,120m,160m,200mFor vertical distanceS=ut+0.5gt2For 1sec, s=45mfor 2 secS=80+20=100mFor 3 secS=120+45=165mFor 4 secS=160+80=240mFor 5secS=200+125=325m
The path of the ball will be parabolic
(b) When the ball hits the ground, its final velocity is zero
H=u22gH=40220=80m
Hope this information will clear your doubts about topic.
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Answer:
What is a Projectile?
What is a Projectile?
Motion Characteristics of a Projectile
Horizontal and Vertical Velocity
Horizontal and Vertical Displacement
Initial Velocity Components
Horizontally Launched Projectile Problems
Non-Horizontally Launched Projectile Problems
In Unit 1 of the Physics Classroom Tutorial, we learned a variety of means to describe the 1-dimensional motion of objects. In Unit 2 of the Physics Classroom Tutorial, we learned how Newton's laws help to explain the motion (and specifically, the changes in the state of motion) of objects that are either at rest or moving in 1-dimension. Now in this unit we will apply both kinematic principles and Newton's laws of motion to understand and explain the motion of objects moving in two dimensions. The most common example of an object that is moving in two dimensions is a projectile. Thus, Lesson 2 of this unit is devoted to understanding the motion of projectiles.