# Acceleration and velocity of projecitile at maximum height, without time

I'm returning to school and taking a physics class. We have a question that is confusing me. An object is thrown up, released from a height of 1m, and reaches a maximum height of 1.5m. Then it asks me to find acceleration and velocity at the maximum height. Don't I need a time component (how long it takes to reach maximum height) to find acceleration and velocity? I thought I was understanding this chapter but apparently not.

• When the ball was going upwards, its velocity was positive. When the ball is coming down, its velocity is negative. Surely, the velocity's sign must have changed at the apex of the projectile. So, what velocity would it have had at the apex? – Gaurang Tandon Jan 27 '18 at 3:08

The problem is meant to help you get better intuition for acceleration and velocity. I'll guess the problem takes place on Earth, so the point here is that the object is always accelerating downwards at $-9.8\mathrm{m}/\mathrm{s}$, from just after it's released to the moment it hits the ground — and even when it's right at the top of its trajectory. Now, as for its velocity, you're asked for the velocity right at the maximum height — where it's not moving up or down. So if the velocity isn't positive or negative, what is it?

Try to remember to think about the physical situation before you start bringing technical terms or math into it. You know this stuff; you just have to trust yourself enough to relate your intuition to the technical side.

• Thanks. This makes some sense. I considered using gravity (9.8 m/s^2) but I was only thinking of it coming down, not as negative acceleration going up. So can I assume it took .051 seconds to reach maximum height? .5/9.8 – morroben Jan 27 '18 at 3:16
• There's really no need to do any math here, because the question (at least as you've stated it) doesn't ask for the time it takes. The acceleration is always $-9.8\mathrm{m}/\mathrm{s}$ as long as the ball is in the air, and the velocity at the top is zero. They gave you that extra information about 1 and 1.5m just to see if you could figure out what's important. Those quantities are not. – Mike Jan 27 '18 at 14:30
• But if you still want to find the time, your expression $.5/9.8$ is not right. Just look at the units you get: seconds$^2$. Your class will probably go over the formula $d = v_0 t + \frac{1}{2}a t^2$, where $d$ is distance, $v_0$ is the initial velocity, $a$ is acceleration, and $t$ is the time it takes. The time it takes to go up is the same as the time it takes to go down. So starting from the top, you have $v_0 = 0$, which means $d = \frac{1}{2}a t^2$. If $d=-0.5\mathrm{m}$ and $a=-9.8\mathrm{m}/\mathrm{s}$, the time it takes is $t = \sqrt{2d/a} = \sqrt{1/9.8}\mathrm{s}$. – Mike Jan 27 '18 at 14:35

Naw, that's enough. This particular question doesn't even require math.

For velocity, if the ball stops moving for a moment at the apex, what's that imply? This is, what's the velocity of something that's not moving?

For acceleration, what forces are acting on the ball? That one's going to be constant - everything being pulled by just gravity is going to have a well-known value for acceleration.

## protected by Qmechanic♦Jan 27 '18 at 7:19

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