Projectile and forces formulae question

AI Thread Summary
The discussion focuses on calculating the time of flight and the initial horizontal velocity of a motorcycle stuntman who takes off from a height of 15.0m and lands 60.0m away. The time of flight was correctly calculated as 1.75 seconds using the vertical motion equation, where the initial vertical velocity is zero. To find the horizontal velocity at take-off, the horizontal distance is divided by the time of flight, resulting in a speed of 34.3 m/s. The key point is that the horizontal and vertical motions are independent; while vertical motion is influenced by gravity, horizontal motion remains constant. Understanding this separation of components is crucial for solving projectile motion problems effectively.
Kurokari
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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 =)
 
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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!
 
Mandeep Deka said:
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?
 
Kurokari said:
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 uyt + (1/2)gt2 equation, in which the initial velocity uy 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 ux.
 
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