Question

a ball is thrown with a speed of 17.7 m/s at an angle of 49.8° above the horizontal. how much time does the ball need to reach a height of 4.7 m above the release point ?​

Answer 1

The ball needs approximately
`0.41\; \rm s` to reach that height for the first time.

Step-by-step explanation:

The initial speed of the ball
`17.7\; \rm m \cdot s^(-1)`. However, what would be the initial vertical speed of this ball?

The angle of elevation is
`\theta = 49.8^\circ`. Consider the initial speed of this ball as the length of the hypotenuse of a right triangle. If the angle of elevation is one of the two acute angles of this triangle, the initial vertical speed of this ball would be the leg opposite to that angle.

`\displaystyle \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}`.

`\displaystyle \sin \theta = \frac{v(\text{vertical, initial})}{v(\text{initial})}`.

Therefore:

`\begin{aligned}&v(\text{vertical, initial})\\ &= v(\text{initial}) \cdot \sin\theta \\ &= 17.7\; \rm m \cdot s^(-1) * \sin \left(49.8^\circ\right) \approx 13.5\; \rm m \cdot s^(-1)\end{aligned}`.

Let
`t` denote the time (in seconds) required for the ball to reach a height of
`4.7\; \rm m`.

Let
`g` denote the acceleration because of gravity (typically
`g \approx 9.81\; \rm m \cdot s^(-2)` near the surface of the earth.) The height of the ball
`t` seconds after it was thrown would be:
`\displaystyle (1)/(2)\, g \cdot t^(2) + v(\text{vertical, initial}) \cdot t`.

Assume that
`g \approx 9.81\; \rm m \cdot s^(-2)`. Set the value of this expression for height to
`4.7\; \rm m` and solve for
`t`:

`\displaystyle (1)/(2) * 9.81 \, t^(2) + 13.5 \, t = 4.7`.

Either
`t \approx 0.41` or
`t \approx 2.3` will satisfy this equation. Both of these two values are reasonable. The first value for
`t` (
`0.41\; \rm s`) is the time required for the ball to reach a height of
`4.7\; \rm m` for the first time. The second value (
`2.3\; \rm s`) is the time required for the ball to come under that height on its way back to the ground. The question seems to be asking only for the first (the smaller one) of these two times.