Question

a girl throws a marshmallow that lands in her friends mouth 92m away. The girl threw the marshmallow at an angle of 67 degrees. How hard was did she throw the marshmallow?

Answer 1

The girl threw the marshmallow with a speed of 35.4 m/s.

Step-by-step explanation:

Use the formula for the range of projectile motion:

`R = (v^2\cdot \sin2\theta)/(g)`

with R the range or distance (92m), theta the elevation angle (67 degrees) and g the gravitational acceleration. The velocity can be determined from this as follows:

`|v| = \sqrt{(R\cdot g)/(\sin 2 \theta)} = \sqrt{(92m\cdot 9.8 (m)/(s^2))/(\sin 134^\circ)}=35.40(m)/(s)`

The girl threw the marshmallow with a speed of 35.4 m/s, which is approximately 79 mph. No word on the boy's condition after the marshmallow's landing.

Answer 2

The answer is 25 miles per hour