Question

A balloon is climbing vertically upwards with a constant velocity of 4.2ms".

a sandbag is dropped from the balloon and hits the ground 3.6 seconds later.
calculate or find:

a) velocity of the sandbag as it hits the ground.

b) height of the sandbag at the instant that it is dropped.

Answer 1

The velocity of the sandbag as it hits the ground is 38.16 m/s (downwards). The height of the sandbag at the instant it is dropped is 0 meters.

Step-by-step explanation:

To find the velocity of the sandbag as it hits the ground, we can use the formula v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time taken. In this case, the initial velocity is 4.2 m/s (upwards) and the time taken is 3.6 seconds. Plugging in these values, we get v = 4.2 + (9.8)(3.6) = 38.16 m/s (downwards).

To find the height of the sandbag at the instant it is dropped, we can use the equation h = ut + (1/2)gt², where h is the height, u is the initial velocity, t is the time taken, and g is the acceleration due to gravity. In this case, the initial velocity is 4.2 m/s (upwards) and the time taken is 0 seconds since it is dropped instantaneously. Plugging in these values, we get h = (4.2)(0) + (1/2)(9.8)(0²) = 0 meters.