Before coming to Earth, your teacher was on planet Krypton where there was almos

Question

Before coming to Earth, your teacher was on planet Krypton where there was almost no atmosphere. Your teacher observed and plotted the position of a projectile at constant time intervals of 0.27 s, as shown.

Calculate the horizontal velocity of the projectile.  Answer in units of m/s Your answer must be within

Before coming to Earth, your teacher was on planet Krypton where there was almost no atmosphere. Your teacher observed and plotted the position of a projectile at constant time intervals of 0.27 s, as shown. Calculate the horizontal velocity of the projectile. Answer in units of m/s Your answer must be within plusminus 3.0% Calculate the vertical acceleration due to the gravity of planet Krypton on the projectile. Answer in units of m/s2. Your answer must be within plusminus 5.0%

Explanation / Answer

What you must remember is that because this is on another planet, the acceleration due to gravity is different.

Now the horizontal velocity always remains constant in projectile motion

Hence distance in horizontal direction = horizontal velocity * time

At t= 0 sec, horizontal dist = 0 m

and at t= 0.27 sec, horizontal dist = 0.5 m

at t= 2 * 0.27 = 0.54 sec, horizontal dist = 1.5 m

This means that in a time interval of 0.27 sec, it has moved 1 m

so horizontal speed = 1/0.27 = 3.7 m/sec (Answer)

You can observe that the answer is no matter which intreval of time we take

2) to solve the second part, remember S = ut +0.5 * at^2

where S is vertical velocity at time t

u is initial velocity and a is acceleration,

Now if we consider the phase when the projectile is going up, then a is negative

Take two time instants t1 and t2 and let the distance be S1 and S2 respectively

then S1 = u * t1 - 0.5 * a * t1^2

S2 = u * t2 - 0.5 * a * t2^2

so S2 - S1 = u * (t2 - t1) - 0.5 * a ( t2^2 - t1^2)

Let s take the points (0.27 sec , 5 m) and (3 * 0.27, 55 m)

Then 50 = u * 0.54 - 0.5 * a * 0.583 = 0.54 u - 0.29 a

Now also see that the projectile reaches the highest point at t = 6.5 * 0.27 seconds

at highest point v = 0

so u = at => u = a * 6.5 * 0.27 = 1.755 a

Replacing this back in our previous equationa

50 = 0.54 u - 0.29 a = 0.54 * 1.755 * a - 0.29 a = 0.947 a - 0.29 a = 0.657 a

so a = 50 / 0.657 = 76.103 m/sec^2

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