Projectile and forces formulae question

[Kurokari]

Homework Statement

A motorcycle stuntman who is moving horizontally takes off from a point 15.0m above the ground and lands 60.0m

Calculate
a. Time between taking off and landing
b. the speed of the motorcycle at take-off

Homework Equations

s = ut + 1/2(at^2)

The Attempt at a Solution

Using the above equation and the information, s= 15m , initial velocity, u = 0 , a = g =9.81
the answer was 1.75s, which according to the book is correct.

My problem lies with the second part of it, it is asking me to look for the initial velocity which is at the point of taking off, but I just used u = 0.

The answer given is,

Consider horizontal motion, v = constant.

Horizontal distance = vt

v = 60.0/1.75 = 34.3ms^-1If anyone can help me solve this contradiction and explain it to me, it would really help me understand concept questions like these. Thanks in advance =)

 

[Mandeep Deka]

See, as the stuntman takes off horizontally, w.r.t the ground there isn't any initial velocity in the vertical direction, so analyzing the motion along that direction, its just like dropping a stone from a height of 15 m, so your equation to determine the time of motion gives you the correct answer.
But, since there isn't any force acting along the horizontal direction (neglecting air drag in an ideal situation), the velocity with which the stuntman takes off (i.e the horizontal velocity of the bike at the take off) will not change, and will be a constant till the landing. So you have to use the range then to calculate the velocity, and since you have calculated the time, using the vertical motion of the bike, you can put it in there to determine the velocity of take off, as suggested in the book!

 

[Kurokari]

See, as the stuntman takes off horizontally, w.r.t the ground there isn't any initial velocity in the vertical direction, so analyzing the motion along that direction, its just like dropping a stone from a height of 15 m, so your equation to determine the time of motion gives you the correct answer.
But, since there isn't any force acting along the horizontal direction (neglecting air drag in an ideal situation), the velocity with which the stuntman takes off (i.e the horizontal velocity of the bike at the take off) will not change, and will be a constant till the landing. So you have to use the range then to calculate the velocity, and since you have calculated the time, using the vertical motion of the bike, you can put it in there to determine the velocity of take off, as suggested in the book!

So you're saying in the first part, we just ignore the horizontal component, and just using the concept, how much increase is in the vertical component varying with time but in actual fact, the bike is moving but only horizontally?

 

[gneill]

So you're saying in the first part, we just ignore the horizontal component, and just using the concept, how much increase is in the vertical component varying with time but in actual fact, the bike is moving but only horizontally?

The horizontal and vertical components of the motion are independent. Over the trajectory the motorcycle drops 15m vertically and travels 60.0m in the horizontal direction. You solved for the time of the "flight" by considering first only the motion in the vertical direction, governed by the u_yt + (1/2)gt² equation, in which the initial velocity u_y in the vertical direction was zero.

With the time to landing calculated, you now must consider the motion in the horizontal direction. This is unaccelerated motion at constant velocity. So, choose an appropriate expression and find u_x.